separable equationsseparable equations ivp’sorthogonal ...apilking/math10560/lectures/lecture...
TRANSCRIPT
![Page 1: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/1.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations
A Separable Differential Equation is a first order differential equation of theform
dy
dx= f (y)g(x)
. In this case we have Z1
f (y)dy =
Zg(x)dx .
We can see this by differentiating both sides with respect to x . vskip .01inWhen we perform the above integration, we get an equation relating x and ywhich defines y implicitly as a function of x . Sometimes we can solve for yexplicitly in terms of x .Example Solve the following differential equation: y ′ =
√x
3y.
I dydx
=√
x3y
I Separating variables, we get 3ydy =√
xdx andR
3ydy =R √
xdx
I 3y2
2= x3/2
3/2+ C
I y 2 = 49x3/2 + C or y = ±
q49x3/2 + C
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 2: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/2.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations
A Separable Differential Equation is a first order differential equation of theform
dy
dx= f (y)g(x)
. In this case we have Z1
f (y)dy =
Zg(x)dx .
We can see this by differentiating both sides with respect to x . vskip .01inWhen we perform the above integration, we get an equation relating x and ywhich defines y implicitly as a function of x . Sometimes we can solve for yexplicitly in terms of x .Example Solve the following differential equation: y ′ =
√x
3y.
I dydx
=√
x3y
I Separating variables, we get 3ydy =√
xdx andR
3ydy =R √
xdx
I 3y2
2= x3/2
3/2+ C
I y 2 = 49x3/2 + C or y = ±
q49x3/2 + C
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 3: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/3.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations
A Separable Differential Equation is a first order differential equation of theform
dy
dx= f (y)g(x)
. In this case we have Z1
f (y)dy =
Zg(x)dx .
We can see this by differentiating both sides with respect to x . vskip .01inWhen we perform the above integration, we get an equation relating x and ywhich defines y implicitly as a function of x . Sometimes we can solve for yexplicitly in terms of x .Example Solve the following differential equation: y ′ =
√x
3y.
I dydx
=√
x3y
I Separating variables, we get 3ydy =√
xdx andR
3ydy =R √
xdx
I 3y2
2= x3/2
3/2+ C
I y 2 = 49x3/2 + C or y = ±
q49x3/2 + C
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 4: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/4.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations
A Separable Differential Equation is a first order differential equation of theform
dy
dx= f (y)g(x)
. In this case we have Z1
f (y)dy =
Zg(x)dx .
We can see this by differentiating both sides with respect to x . vskip .01inWhen we perform the above integration, we get an equation relating x and ywhich defines y implicitly as a function of x . Sometimes we can solve for yexplicitly in terms of x .Example Solve the following differential equation: y ′ =
√x
3y.
I dydx
=√
x3y
I Separating variables, we get 3ydy =√
xdx andR
3ydy =R √
xdx
I 3y2
2= x3/2
3/2+ C
I y 2 = 49x3/2 + C or y = ±
q49x3/2 + C
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 5: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/5.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations
A Separable Differential Equation is a first order differential equation of theform
dy
dx= f (y)g(x)
. In this case we have Z1
f (y)dy =
Zg(x)dx .
We can see this by differentiating both sides with respect to x . vskip .01inWhen we perform the above integration, we get an equation relating x and ywhich defines y implicitly as a function of x . Sometimes we can solve for yexplicitly in terms of x .Example Solve the following differential equation: y ′ =
√x
3y.
I dydx
=√
x3y
I Separating variables, we get 3ydy =√
xdx andR
3ydy =R √
xdx
I 3y2
2= x3/2
3/2+ C
I y 2 = 49x3/2 + C or y = ±
q49x3/2 + C
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 6: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/6.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations
Example Solve the following differential equation:
xy + y ′ = 100x
I We write the equation in the form dydx
= f (y)g(x).
I dydx
= 100x − xy = x(100− y)
I Now we separate the variables:
1
100− ydy = xdx .
I Integrate both sides: Z1
100− ydy =
Zxdx .
I with u = 100− y , we have dy = −dy andR
1100−y
dy = − ln |100− y |+ C .
I Therefore, we get − ln |100− y | = x2
2+ C and solving for y , we get
I |100− y | = e−x2/2e−C → 100− y = ±Ke−x2/2 → y = 100 + Ke−x2/2.(can make the K negative if necessary)
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 7: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/7.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations
Example Solve the following differential equation:
xy + y ′ = 100x
I We write the equation in the form dydx
= f (y)g(x).
I dydx
= 100x − xy = x(100− y)
I Now we separate the variables:
1
100− ydy = xdx .
I Integrate both sides: Z1
100− ydy =
Zxdx .
I with u = 100− y , we have dy = −dy andR
1100−y
dy = − ln |100− y |+ C .
I Therefore, we get − ln |100− y | = x2
2+ C and solving for y , we get
I |100− y | = e−x2/2e−C → 100− y = ±Ke−x2/2 → y = 100 + Ke−x2/2.(can make the K negative if necessary)
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 8: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/8.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations
Example Solve the following differential equation:
xy + y ′ = 100x
I We write the equation in the form dydx
= f (y)g(x).
I dydx
= 100x − xy = x(100− y)
I Now we separate the variables:
1
100− ydy = xdx .
I Integrate both sides: Z1
100− ydy =
Zxdx .
I with u = 100− y , we have dy = −dy andR
1100−y
dy = − ln |100− y |+ C .
I Therefore, we get − ln |100− y | = x2
2+ C and solving for y , we get
I |100− y | = e−x2/2e−C → 100− y = ±Ke−x2/2 → y = 100 + Ke−x2/2.(can make the K negative if necessary)
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 9: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/9.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations
Example Solve the following differential equation:
xy + y ′ = 100x
I We write the equation in the form dydx
= f (y)g(x).
I dydx
= 100x − xy = x(100− y)
I Now we separate the variables:
1
100− ydy = xdx .
I Integrate both sides: Z1
100− ydy =
Zxdx .
I with u = 100− y , we have dy = −dy andR
1100−y
dy = − ln |100− y |+ C .
I Therefore, we get − ln |100− y | = x2
2+ C and solving for y , we get
I |100− y | = e−x2/2e−C → 100− y = ±Ke−x2/2 → y = 100 + Ke−x2/2.(can make the K negative if necessary)
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 10: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/10.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations
Example Solve the following differential equation:
xy + y ′ = 100x
I We write the equation in the form dydx
= f (y)g(x).
I dydx
= 100x − xy = x(100− y)
I Now we separate the variables:
1
100− ydy = xdx .
I Integrate both sides: Z1
100− ydy =
Zxdx .
I with u = 100− y , we have dy = −dy andR
1100−y
dy = − ln |100− y |+ C .
I Therefore, we get − ln |100− y | = x2
2+ C and solving for y , we get
I |100− y | = e−x2/2e−C → 100− y = ±Ke−x2/2 → y = 100 + Ke−x2/2.(can make the K negative if necessary)
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 11: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/11.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations
Example Solve the following differential equation:
xy + y ′ = 100x
I We write the equation in the form dydx
= f (y)g(x).
I dydx
= 100x − xy = x(100− y)
I Now we separate the variables:
1
100− ydy = xdx .
I Integrate both sides: Z1
100− ydy =
Zxdx .
I with u = 100− y , we have dy = −dy andR
1100−y
dy = − ln |100− y |+ C .
I Therefore, we get − ln |100− y | = x2
2+ C and solving for y , we get
I |100− y | = e−x2/2e−C → 100− y = ±Ke−x2/2 → y = 100 + Ke−x2/2.(can make the K negative if necessary)
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 12: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/12.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations
Example Solve the following differential equation:
xy + y ′ = 100x
I We write the equation in the form dydx
= f (y)g(x).
I dydx
= 100x − xy = x(100− y)
I Now we separate the variables:
1
100− ydy = xdx .
I Integrate both sides: Z1
100− ydy =
Zxdx .
I with u = 100− y , we have dy = −dy andR
1100−y
dy = − ln |100− y |+ C .
I Therefore, we get − ln |100− y | = x2
2+ C and solving for y , we get
I |100− y | = e−x2/2e−C → 100− y = ±Ke−x2/2 → y = 100 + Ke−x2/2.(can make the K negative if necessary)
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 13: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/13.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations
Example Solve the following differential equation:
xy + y ′ = 100x
I We write the equation in the form dydx
= f (y)g(x).
I dydx
= 100x − xy = x(100− y)
I Now we separate the variables:
1
100− ydy = xdx .
I Integrate both sides: Z1
100− ydy =
Zxdx .
I with u = 100− y , we have dy = −dy andR
1100−y
dy = − ln |100− y |+ C .
I Therefore, we get − ln |100− y | = x2
2+ C and solving for y , we get
I |100− y | = e−x2/2e−C → 100− y = ±Ke−x2/2 → y = 100 + Ke−x2/2.(can make the K negative if necessary)
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 14: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/14.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations IVP’s
Example Solve the following initial value problem:
y ′ =y(sin x)
y 2 + 1, y(0) = 1
I Separating variables we get y2+1y
dy = sin xdx .
I Integrating, we getR
y + 1ydy =
Rsin xdx .
I y2
2+ ln |y | = − cos x + C
I Using the initial value condition y(0) = 1, we get
1/2 + ln(1) = − cos(0) + C
orC = 3/2
I Hence our solution is
y 2
2+ ln |y | = − cos x + 3/2
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 15: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/15.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations IVP’s
Example Solve the following initial value problem:
y ′ =y(sin x)
y 2 + 1, y(0) = 1
I Separating variables we get y2+1y
dy = sin xdx .
I Integrating, we getR
y + 1ydy =
Rsin xdx .
I y2
2+ ln |y | = − cos x + C
I Using the initial value condition y(0) = 1, we get
1/2 + ln(1) = − cos(0) + C
orC = 3/2
I Hence our solution is
y 2
2+ ln |y | = − cos x + 3/2
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 16: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/16.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations IVP’s
Example Solve the following initial value problem:
y ′ =y(sin x)
y 2 + 1, y(0) = 1
I Separating variables we get y2+1y
dy = sin xdx .
I Integrating, we getR
y + 1ydy =
Rsin xdx .
I y2
2+ ln |y | = − cos x + C
I Using the initial value condition y(0) = 1, we get
1/2 + ln(1) = − cos(0) + C
orC = 3/2
I Hence our solution is
y 2
2+ ln |y | = − cos x + 3/2
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 17: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/17.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations IVP’s
Example Solve the following initial value problem:
y ′ =y(sin x)
y 2 + 1, y(0) = 1
I Separating variables we get y2+1y
dy = sin xdx .
I Integrating, we getR
y + 1ydy =
Rsin xdx .
I y2
2+ ln |y | = − cos x + C
I Using the initial value condition y(0) = 1, we get
1/2 + ln(1) = − cos(0) + C
orC = 3/2
I Hence our solution is
y 2
2+ ln |y | = − cos x + 3/2
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 18: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/18.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations IVP’s
Example Solve the following initial value problem:
y ′ =y(sin x)
y 2 + 1, y(0) = 1
I Separating variables we get y2+1y
dy = sin xdx .
I Integrating, we getR
y + 1ydy =
Rsin xdx .
I y2
2+ ln |y | = − cos x + C
I Using the initial value condition y(0) = 1, we get
1/2 + ln(1) = − cos(0) + C
orC = 3/2
I Hence our solution is
y 2
2+ ln |y | = − cos x + 3/2
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 19: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/19.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations IVP’s
Example Solve the following initial value problem:
y ′ =y(sin x)
y 2 + 1, y(0) = 1
I Separating variables we get y2+1y
dy = sin xdx .
I Integrating, we getR
y + 1ydy =
Rsin xdx .
I y2
2+ ln |y | = − cos x + C
I Using the initial value condition y(0) = 1, we get
1/2 + ln(1) = − cos(0) + C
orC = 3/2
I Hence our solution is
y 2
2+ ln |y | = − cos x + 3/2
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 20: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/20.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations IVP’s
Example Find a solution to the following IVP y ′ = yex , y(0) = 2.
I Separating variables, we get 1ydy = exdx .
I Integrating both sides, we get ln |y | = ex + C .
I y = Keex
I Using the initial value condition y(0) = 2, we get
2 = Ke, or K = 2/e
I Our solution becomes
y =2
eeex
= 2e(ex−1).
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 21: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/21.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations IVP’s
Example Find a solution to the following IVP y ′ = yex , y(0) = 2.
I Separating variables, we get 1ydy = exdx .
I Integrating both sides, we get ln |y | = ex + C .
I y = Keex
I Using the initial value condition y(0) = 2, we get
2 = Ke, or K = 2/e
I Our solution becomes
y =2
eeex
= 2e(ex−1).
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 22: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/22.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations IVP’s
Example Find a solution to the following IVP y ′ = yex , y(0) = 2.
I Separating variables, we get 1ydy = exdx .
I Integrating both sides, we get ln |y | = ex + C .
I y = Keex
I Using the initial value condition y(0) = 2, we get
2 = Ke, or K = 2/e
I Our solution becomes
y =2
eeex
= 2e(ex−1).
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 23: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/23.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations IVP’s
Example Find a solution to the following IVP y ′ = yex , y(0) = 2.
I Separating variables, we get 1ydy = exdx .
I Integrating both sides, we get ln |y | = ex + C .
I y = Keex
I Using the initial value condition y(0) = 2, we get
2 = Ke, or K = 2/e
I Our solution becomes
y =2
eeex
= 2e(ex−1).
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 24: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/24.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations IVP’s
Example Find a solution to the following IVP y ′ = yex , y(0) = 2.
I Separating variables, we get 1ydy = exdx .
I Integrating both sides, we get ln |y | = ex + C .
I y = Keex
I Using the initial value condition y(0) = 2, we get
2 = Ke, or K = 2/e
I Our solution becomes
y =2
eeex
= 2e(ex−1).
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 25: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/25.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Separable Equations IVP’s
Example Find a solution to the following IVP y ′ = yex , y(0) = 2.
I Separating variables, we get 1ydy = exdx .
I Integrating both sides, we get ln |y | = ex + C .
I y = Keex
I Using the initial value condition y(0) = 2, we get
2 = Ke, or K = 2/e
I Our solution becomes
y =2
eeex
= 2e(ex−1).
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 26: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/26.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Orthogonal Trajectories.
An Orthogonal Trajectory of a family of curves is a curve that intersects eachcurve in the family of curves at right angles. Two curves intersect at rightangles if their tangents at that point intersect at right angles. That is if theproduct of their slopes at the point of intersection is −1.
Example The family of curves x2 + y 2 = a2 is a family of concentric circlescentered at the origin. The family of lines of form y = kx , are a family oforthogonal trajectories. Why?
I Geometrically, we know that atangent to a circle isperpendicular to the radial linefrom the point of contact to thecenter of the circle.
I Algebraically; At any point (x , y)on the circle x2 + y 2 = a2, wehave 2x + 2y dy
dx= 0. Therefore,
dydx
= − xy
. At the point (x , kx),
we have dydx
= − 1k
.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 27: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/27.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Orthogonal Trajectories.
An Orthogonal Trajectory of a family of curves is a curve that intersects eachcurve in the family of curves at right angles. Two curves intersect at rightangles if their tangents at that point intersect at right angles. That is if theproduct of their slopes at the point of intersection is −1.
Example The family of curves x2 + y 2 = a2 is a family of concentric circlescentered at the origin. The family of lines of form y = kx , are a family oforthogonal trajectories. Why?
I Geometrically, we know that atangent to a circle isperpendicular to the radial linefrom the point of contact to thecenter of the circle.
I Algebraically; At any point (x , y)on the circle x2 + y 2 = a2, wehave 2x + 2y dy
dx= 0. Therefore,
dydx
= − xy
. At the point (x , kx),
we have dydx
= − 1k
.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 28: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/28.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Orthogonal Trajectories.
An Orthogonal Trajectory of a family of curves is a curve that intersects eachcurve in the family of curves at right angles. Two curves intersect at rightangles if their tangents at that point intersect at right angles. That is if theproduct of their slopes at the point of intersection is −1.
Example The family of curves x2 + y 2 = a2 is a family of concentric circlescentered at the origin. The family of lines of form y = kx , are a family oforthogonal trajectories. Why?
I Geometrically, we know that atangent to a circle isperpendicular to the radial linefrom the point of contact to thecenter of the circle.
I Algebraically; At any point (x , y)on the circle x2 + y 2 = a2, wehave 2x + 2y dy
dx= 0. Therefore,
dydx
= − xy
. At the point (x , kx),
we have dydx
= − 1k
.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 29: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/29.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Orthogonal Trajectories.
To find the orthogonal trajectories to a family of curves,
I We find a differential equation satisfied by all of the curves solving for anyconstants in the description in terms of x and y .
I We then use the fact that the products of the derivatives of orthogonalcurves is −1 to find a new differential equation whose solution is thefamily of orthogonal trajectories.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 30: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/30.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Orthogonal Trajectories.
Example Find the orthogonal trajectories to the family of curves y = ax3.
I We differentiate both sides to get: dydx
= 3ax2.
I We must now eliminate the constant a to find a differential equation interms of x and y . We use the original equation to solve for a in terms ofx and y . We get a = y
x3 .
I Substituting into our differential equation, we get dydx
= 3 yx3 x2 = 3 y
x.
I This differential equation describes the family of curves given above.Their orthogonal trajectories must satisfy the differential equation
dy
dx= − x
3y
because the product of the derivatives must be -1.
I Solving this differential equation gives us the orthogonal trajectories.
I Separating variables, we get 3ydy = −xdx →R
3ydy = −R
xdx .
I Therefore 3 y2
2= − x2
2+ C or y2
2/3+ x2
2= C . This is a family of ellipses.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 31: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/31.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Orthogonal Trajectories.
Example Find the orthogonal trajectories to the family of curves y = ax3.
I We differentiate both sides to get: dydx
= 3ax2.
I We must now eliminate the constant a to find a differential equation interms of x and y . We use the original equation to solve for a in terms ofx and y . We get a = y
x3 .
I Substituting into our differential equation, we get dydx
= 3 yx3 x2 = 3 y
x.
I This differential equation describes the family of curves given above.Their orthogonal trajectories must satisfy the differential equation
dy
dx= − x
3y
because the product of the derivatives must be -1.
I Solving this differential equation gives us the orthogonal trajectories.
I Separating variables, we get 3ydy = −xdx →R
3ydy = −R
xdx .
I Therefore 3 y2
2= − x2
2+ C or y2
2/3+ x2
2= C . This is a family of ellipses.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 32: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/32.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Orthogonal Trajectories.
Example Find the orthogonal trajectories to the family of curves y = ax3.
I We differentiate both sides to get: dydx
= 3ax2.
I We must now eliminate the constant a to find a differential equation interms of x and y . We use the original equation to solve for a in terms ofx and y . We get a = y
x3 .
I Substituting into our differential equation, we get dydx
= 3 yx3 x2 = 3 y
x.
I This differential equation describes the family of curves given above.Their orthogonal trajectories must satisfy the differential equation
dy
dx= − x
3y
because the product of the derivatives must be -1.
I Solving this differential equation gives us the orthogonal trajectories.
I Separating variables, we get 3ydy = −xdx →R
3ydy = −R
xdx .
I Therefore 3 y2
2= − x2
2+ C or y2
2/3+ x2
2= C . This is a family of ellipses.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 33: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/33.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Orthogonal Trajectories.
Example Find the orthogonal trajectories to the family of curves y = ax3.
I We differentiate both sides to get: dydx
= 3ax2.
I We must now eliminate the constant a to find a differential equation interms of x and y . We use the original equation to solve for a in terms ofx and y . We get a = y
x3 .
I Substituting into our differential equation, we get dydx
= 3 yx3 x2 = 3 y
x.
I This differential equation describes the family of curves given above.Their orthogonal trajectories must satisfy the differential equation
dy
dx= − x
3y
because the product of the derivatives must be -1.
I Solving this differential equation gives us the orthogonal trajectories.
I Separating variables, we get 3ydy = −xdx →R
3ydy = −R
xdx .
I Therefore 3 y2
2= − x2
2+ C or y2
2/3+ x2
2= C . This is a family of ellipses.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 34: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/34.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Orthogonal Trajectories.
Example Find the orthogonal trajectories to the family of curves y = ax3.
I We differentiate both sides to get: dydx
= 3ax2.
I We must now eliminate the constant a to find a differential equation interms of x and y . We use the original equation to solve for a in terms ofx and y . We get a = y
x3 .
I Substituting into our differential equation, we get dydx
= 3 yx3 x2 = 3 y
x.
I This differential equation describes the family of curves given above.Their orthogonal trajectories must satisfy the differential equation
dy
dx= − x
3y
because the product of the derivatives must be -1.
I Solving this differential equation gives us the orthogonal trajectories.
I Separating variables, we get 3ydy = −xdx →R
3ydy = −R
xdx .
I Therefore 3 y2
2= − x2
2+ C or y2
2/3+ x2
2= C . This is a family of ellipses.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 35: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/35.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Orthogonal Trajectories.
Example Find the orthogonal trajectories to the family of curves y = ax3.
I We differentiate both sides to get: dydx
= 3ax2.
I We must now eliminate the constant a to find a differential equation interms of x and y . We use the original equation to solve for a in terms ofx and y . We get a = y
x3 .
I Substituting into our differential equation, we get dydx
= 3 yx3 x2 = 3 y
x.
I This differential equation describes the family of curves given above.Their orthogonal trajectories must satisfy the differential equation
dy
dx= − x
3y
because the product of the derivatives must be -1.
I Solving this differential equation gives us the orthogonal trajectories.
I Separating variables, we get 3ydy = −xdx →R
3ydy = −R
xdx .
I Therefore 3 y2
2= − x2
2+ C or y2
2/3+ x2
2= C . This is a family of ellipses.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 36: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/36.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Orthogonal Trajectories.
Example Find the orthogonal trajectories to the family of curves y = ax3.
I We differentiate both sides to get: dydx
= 3ax2.
I We must now eliminate the constant a to find a differential equation interms of x and y . We use the original equation to solve for a in terms ofx and y . We get a = y
x3 .
I Substituting into our differential equation, we get dydx
= 3 yx3 x2 = 3 y
x.
I This differential equation describes the family of curves given above.Their orthogonal trajectories must satisfy the differential equation
dy
dx= − x
3y
because the product of the derivatives must be -1.
I Solving this differential equation gives us the orthogonal trajectories.
I Separating variables, we get 3ydy = −xdx →R
3ydy = −R
xdx .
I Therefore 3 y2
2= − x2
2+ C or y2
2/3+ x2
2= C . This is a family of ellipses.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 37: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/37.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Orthogonal Trajectories.
Example Find the orthogonal trajectories to the family of curves y = ax3.
I We differentiate both sides to get: dydx
= 3ax2.
I We must now eliminate the constant a to find a differential equation interms of x and y . We use the original equation to solve for a in terms ofx and y . We get a = y
x3 .
I Substituting into our differential equation, we get dydx
= 3 yx3 x2 = 3 y
x.
I This differential equation describes the family of curves given above.Their orthogonal trajectories must satisfy the differential equation
dy
dx= − x
3y
because the product of the derivatives must be -1.
I Solving this differential equation gives us the orthogonal trajectories.
I Separating variables, we get 3ydy = −xdx →R
3ydy = −R
xdx .
I Therefore 3 y2
2= − x2
2+ C or y2
2/3+ x2
2= C . This is a family of ellipses.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 38: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/38.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Orthogonal Trajectories.
Example Find the orthogonal trajectories to the family of curves y = ax3.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 39: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/39.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
In these problems a chemical in a liquid solution (or gas) runs into a containerholding the liquid. The liquid in the container may already have a specifiedamount of the chemical dissolved in it. We assume the mixture is kept uniformby stirring and flows out of the container at a known rate. The differentialequation describing the process is based on the formula
Rate of Change Rate at which Rate at whichof the amount = chemical - chemical
in the container arrives departs
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) be the amount of alcohol in the container at time t. Let V (t)denote the amount of beer in the tank at time t.
I In this case V (t) remains constant at V (t) = 500. y(0) = 15 gallons.
I We have the rate of change in the amount of alcohol in the vat is
dy
dt= 5% of 5 gal/min− y(t)
V5 gal per min.
= .05× 5− y
500× 5 = .25− y/100 =
25− y
100.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 40: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/40.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
In these problems a chemical in a liquid solution (or gas) runs into a containerholding the liquid. The liquid in the container may already have a specifiedamount of the chemical dissolved in it. We assume the mixture is kept uniformby stirring and flows out of the container at a known rate. The differentialequation describing the process is based on the formula
Rate of Change Rate at which Rate at whichof the amount = chemical - chemical
in the container arrives departs
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) be the amount of alcohol in the container at time t. Let V (t)denote the amount of beer in the tank at time t.
I In this case V (t) remains constant at V (t) = 500. y(0) = 15 gallons.
I We have the rate of change in the amount of alcohol in the vat is
dy
dt= 5% of 5 gal/min− y(t)
V5 gal per min.
= .05× 5− y
500× 5 = .25− y/100 =
25− y
100.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 41: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/41.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
In these problems a chemical in a liquid solution (or gas) runs into a containerholding the liquid. The liquid in the container may already have a specifiedamount of the chemical dissolved in it. We assume the mixture is kept uniformby stirring and flows out of the container at a known rate. The differentialequation describing the process is based on the formula
Rate of Change Rate at which Rate at whichof the amount = chemical - chemical
in the container arrives departs
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) be the amount of alcohol in the container at time t. Let V (t)denote the amount of beer in the tank at time t.
I In this case V (t) remains constant at V (t) = 500. y(0) = 15 gallons.
I We have the rate of change in the amount of alcohol in the vat is
dy
dt= 5% of 5 gal/min− y(t)
V5 gal per min.
= .05× 5− y
500× 5 = .25− y/100 =
25− y
100.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 42: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/42.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
In these problems a chemical in a liquid solution (or gas) runs into a containerholding the liquid. The liquid in the container may already have a specifiedamount of the chemical dissolved in it. We assume the mixture is kept uniformby stirring and flows out of the container at a known rate. The differentialequation describing the process is based on the formula
Rate of Change Rate at which Rate at whichof the amount = chemical - chemical
in the container arrives departs
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) be the amount of alcohol in the container at time t. Let V (t)denote the amount of beer in the tank at time t.
I In this case V (t) remains constant at V (t) = 500. y(0) = 15 gallons.
I We have the rate of change in the amount of alcohol in the vat is
dy
dt= 5% of 5 gal/min− y(t)
V5 gal per min.
= .05× 5− y
500× 5 = .25− y/100 =
25− y
100.Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 43: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/43.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
Rate of Change Rate at which Rate at whichof the amount = chemical - chemical
in the container arrives departs
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) be the amount of alcohol in the container at time t. Let V (t)denote the amount of beer in the tank at time t.
I In this case V (t) remains constant at V (t) = 500. y(0) = 15 gallons.
I We have the rate of change in the amount of alcohol in the vat is
dy
dt= 5% of 5 gal/min−y(t)
V5 gal per min. = .05×5− y
500×5 = .25−y/100
=25− y
100→ dy
dt=
25− y
100.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 44: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/44.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) = the amount of alcohol in the container at time t. V (t) = theamount of beer in the tank at time t.
I V (t) = 500. y(0) = 15 gallons.
I We have the rate of change in the amount of alcohol in the vat is
dy
dt=
25− y
100.
I We solve this IVP and find y(60).
I Separating variables, we get dy25−y
= 1100
dt.
I Integrating both sides, we get − ln |25− y | = t100
+ C .
I taking the exponential of both sides of ln |25− y | = − t100
+ C we get
|25− y | = eCe−t100 or 25− y = ±eCe
−t100 → 25− y = Ke
−t100 →
y = 25− Ke−t100 .
I Using the initial value condition, we get 15 = 25− Ke0 = 25− K . Hence
K = 10 and y = 25− 10e−t100 .
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 45: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/45.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) = the amount of alcohol in the container at time t. V (t) = theamount of beer in the tank at time t.
I V (t) = 500. y(0) = 15 gallons.
I We have the rate of change in the amount of alcohol in the vat is
dy
dt=
25− y
100.
I We solve this IVP and find y(60).
I Separating variables, we get dy25−y
= 1100
dt.
I Integrating both sides, we get − ln |25− y | = t100
+ C .
I taking the exponential of both sides of ln |25− y | = − t100
+ C we get
|25− y | = eCe−t100 or 25− y = ±eCe
−t100 → 25− y = Ke
−t100 →
y = 25− Ke−t100 .
I Using the initial value condition, we get 15 = 25− Ke0 = 25− K . Hence
K = 10 and y = 25− 10e−t100 .
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 46: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/46.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) = the amount of alcohol in the container at time t. V (t) = theamount of beer in the tank at time t.
I V (t) = 500. y(0) = 15 gallons.
I We have the rate of change in the amount of alcohol in the vat is
dy
dt=
25− y
100.
I We solve this IVP and find y(60).
I Separating variables, we get dy25−y
= 1100
dt.
I Integrating both sides, we get − ln |25− y | = t100
+ C .
I taking the exponential of both sides of ln |25− y | = − t100
+ C we get
|25− y | = eCe−t100 or 25− y = ±eCe
−t100 → 25− y = Ke
−t100 →
y = 25− Ke−t100 .
I Using the initial value condition, we get 15 = 25− Ke0 = 25− K . Hence
K = 10 and y = 25− 10e−t100 .
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 47: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/47.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) = the amount of alcohol in the container at time t. V (t) = theamount of beer in the tank at time t.
I V (t) = 500. y(0) = 15 gallons.
I We have the rate of change in the amount of alcohol in the vat is
dy
dt=
25− y
100.
I We solve this IVP and find y(60).
I Separating variables, we get dy25−y
= 1100
dt.
I Integrating both sides, we get − ln |25− y | = t100
+ C .
I taking the exponential of both sides of ln |25− y | = − t100
+ C we get
|25− y | = eCe−t100 or 25− y = ±eCe
−t100 → 25− y = Ke
−t100 →
y = 25− Ke−t100 .
I Using the initial value condition, we get 15 = 25− Ke0 = 25− K . Hence
K = 10 and y = 25− 10e−t100 .
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 48: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/48.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) = the amount of alcohol in the container at time t. V (t) = theamount of beer in the tank at time t.
I V (t) = 500. y(0) = 15 gallons.
I We have the rate of change in the amount of alcohol in the vat is
dy
dt=
25− y
100.
I We solve this IVP and find y(60).
I Separating variables, we get dy25−y
= 1100
dt.
I Integrating both sides, we get − ln |25− y | = t100
+ C .
I taking the exponential of both sides of ln |25− y | = − t100
+ C we get
|25− y | = eCe−t100 or 25− y = ±eCe
−t100 → 25− y = Ke
−t100 →
y = 25− Ke−t100 .
I Using the initial value condition, we get 15 = 25− Ke0 = 25− K . Hence
K = 10 and y = 25− 10e−t100 .
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 49: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/49.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) = the amount of alcohol in the container at time t. V (t) = theamount of beer in the tank at time t.
I V (t) = 500. y(0) = 15 gallons.
I We have the rate of change in the amount of alcohol in the vat is
dy
dt=
25− y
100.
I We solve this IVP and find y(60).
I Separating variables, we get dy25−y
= 1100
dt.
I Integrating both sides, we get − ln |25− y | = t100
+ C .
I taking the exponential of both sides of ln |25− y | = − t100
+ C we get
|25− y | = eCe−t100 or 25− y = ±eCe
−t100 → 25− y = Ke
−t100 →
y = 25− Ke−t100 .
I Using the initial value condition, we get 15 = 25− Ke0 = 25− K . Hence
K = 10 and y = 25− 10e−t100 .Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 50: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/50.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) = the amount of alcohol in the container at time t. V (t) = theamount of beer in the tank at time t.
I V (t) = 500. y(0) = 15 gallons.
I Using the initial value condition, we get 15 = 25− Ke0 = 25− K . Hence
K = 10 and y = 25− 10e−t100 .
I y(60) = 25− 10e−0.6 = 19.512.
I The percentage of alcohol in the tank after 60 minutes isy(60)500× 100% = 3.9%.
I Note we would expect an answer somewhere between the initial 3% andthe 5% in the beer flowing into the tank.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 51: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/51.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) = the amount of alcohol in the container at time t. V (t) = theamount of beer in the tank at time t.
I V (t) = 500. y(0) = 15 gallons.
I Using the initial value condition, we get 15 = 25− Ke0 = 25− K . Hence
K = 10 and y = 25− 10e−t100 .
I y(60) = 25− 10e−0.6 = 19.512.
I The percentage of alcohol in the tank after 60 minutes isy(60)500× 100% = 3.9%.
I Note we would expect an answer somewhere between the initial 3% andthe 5% in the beer flowing into the tank.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 52: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/52.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) = the amount of alcohol in the container at time t. V (t) = theamount of beer in the tank at time t.
I V (t) = 500. y(0) = 15 gallons.
I Using the initial value condition, we get 15 = 25− Ke0 = 25− K . Hence
K = 10 and y = 25− 10e−t100 .
I y(60) = 25− 10e−0.6 = 19.512.
I The percentage of alcohol in the tank after 60 minutes isy(60)500× 100% = 3.9%.
I Note we would expect an answer somewhere between the initial 3% andthe 5% in the beer flowing into the tank.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method
![Page 53: Separable EquationsSeparable Equations IVP’sOrthogonal ...apilking/Math10560/Lectures/Lecture 19.pdfSeparable EquationsSeparable Equations IVP’sOrthogonal Trajectories.Mixture](https://reader030.vdocument.in/reader030/viewer/2022040820/5e6967528e169714ef53c1d4/html5/thumbnails/53.jpg)
Separable Equations Separable Equations IVP’s Orthogonal Trajectories. Mixture Problems.
Mixture Problems.
Example A vat at Guinness’ brewery with 500 gallons of beer contains 3%alcohol (15 gallons). Beer with 5% alcohol per unit of volume is pumped intothe vat at a rate of 5 gallons/ min and the mixture is pumped out at the samerate. What is the percentage of alcohol after one hour?
I Let y(t) = the amount of alcohol in the container at time t. V (t) = theamount of beer in the tank at time t.
I V (t) = 500. y(0) = 15 gallons.
I Using the initial value condition, we get 15 = 25− Ke0 = 25− K . Hence
K = 10 and y = 25− 10e−t100 .
I y(60) = 25− 10e−0.6 = 19.512.
I The percentage of alcohol in the tank after 60 minutes isy(60)500× 100% = 3.9%.
I Note we would expect an answer somewhere between the initial 3% andthe 5% in the beer flowing into the tank.
Annette Pilkington Lecture 19 : Direction Fields and Euler’s Method