MA 108 - Ordinary Differential Equations

Santanu Dey

Department of Mathematics,Indian Institute of Technology Bombay,

Powai, Mumbai 76dey@math.iitb.ac.in

October 10, 2013

Santanu Dey Lecture 7

Outline of the lecture

Cauchy-Euler equations

Non-homogeneous equations

Santanu Dey Lecture 7

Cauchy-Euler Equations

The equation

x2y ′′ + axy ′ + by = 0

where a, b ∈ R is called a Cauchy-Euler equation . Assume x > 0.

Suppose y = xm is a solution to this DE. Then,

x2m(m − 1)xm−2 + axmxm−1 + bxm = 0.

We get:

m(m − 1) + am + b = 0.

that is,m2 + (a− 1)m + b = 0.

This is called the auxiliary equation of the given Cauchy-Euler equation.The roots are

m1,m2 =(1− a)±

√(a− 1)2 − 4b

2.

Santanu Dey Lecture 7

Cauchy-Euler Equations

The equation

x2y ′′ + axy ′ + by = 0

where a, b ∈ R is called a Cauchy-Euler equation .

Assume x > 0.

Suppose y = xm is a solution to this DE. Then,

x2m(m − 1)xm−2 + axmxm−1 + bxm = 0.

We get:

m(m − 1) + am + b = 0.

that is,m2 + (a− 1)m + b = 0.

This is called the auxiliary equation of the given Cauchy-Euler equation.The roots are

m1,m2 =(1− a)±

√(a− 1)2 − 4b

2.

Santanu Dey Lecture 7

Cauchy-Euler Equations

The equation

x2y ′′ + axy ′ + by = 0

where a, b ∈ R is called a Cauchy-Euler equation . Assume x > 0.

Suppose y = xm is a solution to this DE. Then,

x2m(m − 1)xm−2 + axmxm−1 + bxm = 0.

We get:

m(m − 1) + am + b = 0.

that is,m2 + (a− 1)m + b = 0.

This is called the auxiliary equation of the given Cauchy-Euler equation.The roots are

m1,m2 =(1− a)±

√(a− 1)2 − 4b

2.

Santanu Dey Lecture 7

Cauchy-Euler Equations

The equation

x2y ′′ + axy ′ + by = 0

where a, b ∈ R is called a Cauchy-Euler equation . Assume x > 0.

Suppose y = xm is a solution to this DE.

Then,

x2m(m − 1)xm−2 + axmxm−1 + bxm = 0.

We get:

m(m − 1) + am + b = 0.

that is,m2 + (a− 1)m + b = 0.

This is called the auxiliary equation of the given Cauchy-Euler equation.The roots are

m1,m2 =(1− a)±

√(a− 1)2 − 4b

2.

Santanu Dey Lecture 7

Cauchy-Euler Equations

The equation

x2y ′′ + axy ′ + by = 0

where a, b ∈ R is called a Cauchy-Euler equation . Assume x > 0.

Suppose y = xm is a solution to this DE. Then,

x2m(m − 1)xm−2 + axmxm−1 + bxm = 0.

We get:

m(m − 1) + am + b = 0.

that is,m2 + (a− 1)m + b = 0.

This is called the auxiliary equation of the given Cauchy-Euler equation.The roots are

m1,m2 =(1− a)±

√(a− 1)2 − 4b

2.

Santanu Dey Lecture 7

Cauchy-Euler Equations

The equation

x2y ′′ + axy ′ + by = 0

where a, b ∈ R is called a Cauchy-Euler equation . Assume x > 0.

Suppose y = xm is a solution to this DE. Then,

x2m(m − 1)xm−2 + axmxm−1 + bxm = 0.

We get:

m(m − 1) + am + b = 0.

that is,m2 + (a− 1)m + b = 0.

This is called the auxiliary equation of the given Cauchy-Euler equation.The roots are

m1,m2 =(1− a)±

√(a− 1)2 − 4b

2.

Santanu Dey Lecture 7

Cauchy-Euler Equations

The equation

x2y ′′ + axy ′ + by = 0

where a, b ∈ R is called a Cauchy-Euler equation . Assume x > 0.

Suppose y = xm is a solution to this DE. Then,

x2m(m − 1)xm−2 + axmxm−1 + bxm = 0.

We get:

m(m − 1) + am + b = 0.

that is,m2 + (a− 1)m + b = 0.

This is called the auxiliary equation of the given Cauchy-Euler equation.The roots are

m1,m2 =(1− a)±

√(a− 1)2 − 4b

2.

Santanu Dey Lecture 7

Cauchy-Euler Equations

The equation

x2y ′′ + axy ′ + by = 0

where a, b ∈ R is called a Cauchy-Euler equation . Assume x > 0.

Suppose y = xm is a solution to this DE. Then,

x2m(m − 1)xm−2 + axmxm−1 + bxm = 0.

We get:

m(m − 1) + am + b = 0.

that is,m2 + (a− 1)m + b = 0.

This is called the auxiliary equation of the given Cauchy-Euler equation.The roots are

m1,m2 =(1− a)±

√(a− 1)2 − 4b

2.

Santanu Dey Lecture 7

Cauchy-Euler Equations

The equation

x2y ′′ + axy ′ + by = 0

where a, b ∈ R is called a Cauchy-Euler equation . Assume x > 0.

Suppose y = xm is a solution to this DE. Then,

x2m(m − 1)xm−2 + axmxm−1 + bxm = 0.

We get:

m(m − 1) + am + b = 0.

that is,m2 + (a− 1)m + b = 0.

This is called the auxiliary equation of the given Cauchy-Euler equation.

The roots are

m1,m2 =(1− a)±

√(a− 1)2 − 4b

2.

Santanu Dey Lecture 7

Cauchy-Euler Equations

The equation

x2y ′′ + axy ′ + by = 0

where a, b ∈ R is called a Cauchy-Euler equation . Assume x > 0.

Suppose y = xm is a solution to this DE. Then,

x2m(m − 1)xm−2 + axmxm−1 + bxm = 0.

We get:

m(m − 1) + am + b = 0.

that is,m2 + (a− 1)m + b = 0.

This is called the auxiliary equation of the given Cauchy-Euler equation.The roots are

m1,m2 =(1− a)±

√(a− 1)2 − 4b

2.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case I: Distinct real roots.

Are xm1 and xm2 linearly independent? Yes. Hence the generalsolution is given by

y = c1xm1 + c2x

m2 ,

for c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case I: Distinct real roots.Are xm1 and xm2 linearly independent?

Yes. Hence the generalsolution is given by

y = c1xm1 + c2x

m2 ,

for c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case I: Distinct real roots.Are xm1 and xm2 linearly independent? Yes.

Hence the generalsolution is given by

y = c1xm1 + c2x

m2 ,

for c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case I: Distinct real roots.Are xm1 and xm2 linearly independent? Yes. Hence the generalsolution is given by

y = c1xm1 + c2x

m2 ,

for c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case I: Distinct real roots.Are xm1 and xm2 linearly independent? Yes. Hence the generalsolution is given by

y = c1xm1 + c2x

m2 ,

for c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case I: Distinct real roots.Are xm1 and xm2 linearly independent? Yes. Hence the generalsolution is given by

y = c1xm1 + c2x

m2 ,

for c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case II: Equal real roots.

that is,

m1 = m2 =1− a

2.

Hencey = f (x) = x

1−a2

is a solution. To get a solution g(x) linearly independent fromf (x), set

g(x) = v(x)f (x).

v(x) =

∫e−

∫axdx

x1−adx =

∫dx

x= ln x

Hence,g(x) = (ln x)x

1−a2 .

Thus the general solution is given by

y = c1x1−a2 + c2x

1−a2 ln x ,

c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case II: Equal real roots.that is,

m1 = m2 =1− a

2.

Hencey = f (x) = x

1−a2

is a solution. To get a solution g(x) linearly independent fromf (x), set

g(x) = v(x)f (x).

v(x) =

∫e−

∫axdx

x1−adx =

∫dx

x= ln x

Hence,g(x) = (ln x)x

1−a2 .

Thus the general solution is given by

y = c1x1−a2 + c2x

1−a2 ln x ,

c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case II: Equal real roots.that is,

m1 = m2 =1− a

2.

Hencey = f (x) = x

1−a2

is a solution.

To get a solution g(x) linearly independent fromf (x), set

g(x) = v(x)f (x).

v(x) =

∫e−

∫axdx

x1−adx =

∫dx

x= ln x

Hence,g(x) = (ln x)x

1−a2 .

Thus the general solution is given by

y = c1x1−a2 + c2x

1−a2 ln x ,

c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case II: Equal real roots.that is,

m1 = m2 =1− a

2.

Hencey = f (x) = x

1−a2

is a solution. To get a solution g(x) linearly independent fromf (x),

setg(x) = v(x)f (x).

v(x) =

∫e−

∫axdx

x1−adx =

∫dx

x= ln x

Hence,g(x) = (ln x)x

1−a2 .

Thus the general solution is given by

y = c1x1−a2 + c2x

1−a2 ln x ,

c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case II: Equal real roots.that is,

m1 = m2 =1− a

2.

Hencey = f (x) = x

1−a2

is a solution. To get a solution g(x) linearly independent fromf (x), set

g(x) = v(x)f (x).

v(x) =

∫e−

∫axdx

x1−adx =

∫dx

x= ln x

Hence,g(x) = (ln x)x

1−a2 .

Thus the general solution is given by

y = c1x1−a2 + c2x

1−a2 ln x ,

c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case II: Equal real roots.that is,

m1 = m2 =1− a

2.

Hencey = f (x) = x

1−a2

is a solution. To get a solution g(x) linearly independent fromf (x), set

g(x) = v(x)f (x).

v(x) =

∫e−

∫axdx

x1−adx =

∫dx

x= ln x

Hence,g(x) = (ln x)x

1−a2 .

Thus the general solution is given by

y = c1x1−a2 + c2x

1−a2 ln x ,

c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case II: Equal real roots.that is,

m1 = m2 =1− a

2.

Hencey = f (x) = x

1−a2

is a solution. To get a solution g(x) linearly independent fromf (x), set

g(x) = v(x)f (x).

v(x) =

∫e−

∫axdx

x1−adx =

∫dx

x= ln x

Hence,g(x) = (ln x)x

1−a2 .

Thus the general solution is given by

y = c1x1−a2 + c2x

1−a2 ln x ,

c1, c2 ∈ R. Santanu Dey Lecture 7

Cauchy-Euler Equations

Case III : Complex roots

Roots are m1 = µ+ ıν, m2 = µ− ıν.

xm1 = xµeıν ln x = xµ(cos(ν ln x) + ı sin(ν ln x)),

xm2 = xµe−ıν ln x = xµ(cos(ν ln x)− ı sin(ν ln x)),

General solution is given by

y = xµ(c1 cos(ν ln x) + c2 sin(ν ln x)),

c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case III : Complex roots

Roots are m1 = µ+ ıν, m2 = µ− ıν.

xm1 = xµeıν ln x = xµ(cos(ν ln x) + ı sin(ν ln x)),

xm2 = xµe−ıν ln x = xµ(cos(ν ln x)− ı sin(ν ln x)),

General solution is given by

y = xµ(c1 cos(ν ln x) + c2 sin(ν ln x)),

c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case III : Complex roots

Roots are m1 = µ+ ıν, m2 = µ− ıν.

xm1 = xµeıν ln x = xµ(cos(ν ln x) + ı sin(ν ln x)),

xm2 = xµe−ıν ln x = xµ(cos(ν ln x)− ı sin(ν ln x)),

General solution is given by

y = xµ(c1 cos(ν ln x) + c2 sin(ν ln x)),

c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case III : Complex roots

Roots are m1 = µ+ ıν, m2 = µ− ıν.

xm1 = xµeıν ln x = xµ(cos(ν ln x) + ı sin(ν ln x)),

xm2 = xµe−ıν ln x = xµ(cos(ν ln x)− ı sin(ν ln x)),

General solution is given by

y = xµ(c1 cos(ν ln x) + c2 sin(ν ln x)),

c1, c2 ∈ R.

Santanu Dey Lecture 7

Cauchy-Euler Equations

Case III : Complex roots

Roots are m1 = µ+ ıν, m2 = µ− ıν.

xm1 = xµeıν ln x = xµ(cos(ν ln x) + ı sin(ν ln x)),

xm2 = xµe−ıν ln x = xµ(cos(ν ln x)− ı sin(ν ln x)),

General solution is given by

y = xµ(c1 cos(ν ln x) + c2 sin(ν ln x)),

c1, c2 ∈ R.

Santanu Dey Lecture 7

Examples

Solve:

1 2x2y ′′ + 3xy ′ − y = 0, x > 0.

2 x2y ′′ + 5xy ′ + 4y = 0, x > 0.

3 x2y ′′ + xy ′ + y = 0, x > 0.

Solutions :

1 y = c1√x + c2/x

2 y = x−2(c1 + c2 ln x).

3 y = c1 cos(ln x) + c2 sin(ln x).

Santanu Dey Lecture 7

Examples

Solve:

1 2x2y ′′ + 3xy ′ − y = 0, x > 0.

2 x2y ′′ + 5xy ′ + 4y = 0, x > 0.

3 x2y ′′ + xy ′ + y = 0, x > 0.

Solutions :

1 y = c1√x + c2/x

2 y = x−2(c1 + c2 ln x).

3 y = c1 cos(ln x) + c2 sin(ln x).

Santanu Dey Lecture 7

Examples

Solve:

1 2x2y ′′ + 3xy ′ − y = 0, x > 0.

2 x2y ′′ + 5xy ′ + 4y = 0, x > 0.

3 x2y ′′ + xy ′ + y = 0, x > 0.

Solutions :

1 y = c1√x + c2/x

2 y = x−2(c1 + c2 ln x).

3 y = c1 cos(ln x) + c2 sin(ln x).

Santanu Dey Lecture 7

Examples

Solve:

1 2x2y ′′ + 3xy ′ − y = 0, x > 0.

2 x2y ′′ + 5xy ′ + 4y = 0, x > 0.

3 x2y ′′ + xy ′ + y = 0, x > 0.

Solutions :

1 y = c1√x + c2/x

2 y = x−2(c1 + c2 ln x).

3 y = c1 cos(ln x) + c2 sin(ln x).

Santanu Dey Lecture 7

Examples

Solve:

1 2x2y ′′ + 3xy ′ − y = 0, x > 0.

2 x2y ′′ + 5xy ′ + 4y = 0, x > 0.

3 x2y ′′ + xy ′ + y = 0, x > 0.

Solutions :

1 y = c1√x + c2/x

2 y = x−2(c1 + c2 ln x).

3 y = c1 cos(ln x) + c2 sin(ln x).

Santanu Dey Lecture 7

Examples

Solve:

1 2x2y ′′ + 3xy ′ − y = 0, x > 0.

2 x2y ′′ + 5xy ′ + 4y = 0, x > 0.

3 x2y ′′ + xy ′ + y = 0, x > 0.

Solutions :

1 y = c1√x + c2/x

2 y = x−2(c1 + c2 ln x).

3 y = c1 cos(ln x) + c2 sin(ln x).

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Consider the non-homogeneous DE

y ′′ + p(x)y ′ + q(x)y = r(x)

where p(x), q(x), r(x) are continuous functions on an interval I .

The associated homogeneous DE is

y ′′ + p(x)y ′ + q(x)y = 0.

Can we relate the solutions of the above two DE’s?

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Consider the non-homogeneous DE

y ′′ + p(x)y ′ + q(x)y = r(x)

where p(x), q(x), r(x) are continuous functions on an interval I .The associated homogeneous DE is

y ′′ + p(x)y ′ + q(x)y = 0.

Can we relate the solutions of the above two DE’s?

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Consider the non-homogeneous DE

y ′′ + p(x)y ′ + q(x)y = r(x)

where p(x), q(x), r(x) are continuous functions on an interval I .The associated homogeneous DE is

y ′′ + p(x)y ′ + q(x)y = 0.

Can we relate the solutions of the above two DE’s?

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Theorem

Let yp(x) be any solution of

y ′′ + p(x)y ′ + q(x)y = r(x)

and y1(x), y2(x) be a basis of the solution space of thecorresponding homogeneous DE.Then the set of solutions of the non-homogeneous DE is

{c1y1(x) + c2y2(x) + yp(x) | c1, c2 ∈ R}.

Proof: Let φ(x) be any solution of

L(y) = y ′′ + p(x)y ′ + q(x)y = r(x).

Then,

L(φ(x)− yp(x)) = L(φ(x))− L(yp(x)) = r(x)− r(x) = 0.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Theorem

Let yp(x) be any solution of

y ′′ + p(x)y ′ + q(x)y = r(x)

and y1(x), y2(x) be a basis of the solution space of thecorresponding homogeneous DE.

Then the set of solutions of the non-homogeneous DE is

{c1y1(x) + c2y2(x) + yp(x) | c1, c2 ∈ R}.

Proof: Let φ(x) be any solution of

L(y) = y ′′ + p(x)y ′ + q(x)y = r(x).

Then,

L(φ(x)− yp(x)) = L(φ(x))− L(yp(x)) = r(x)− r(x) = 0.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Theorem

Let yp(x) be any solution of

y ′′ + p(x)y ′ + q(x)y = r(x)

and y1(x), y2(x) be a basis of the solution space of thecorresponding homogeneous DE.Then the set of solutions of the non-homogeneous DE is

{c1y1(x) + c2y2(x) + yp(x) | c1, c2 ∈ R}.

Proof: Let φ(x) be any solution of

L(y) = y ′′ + p(x)y ′ + q(x)y = r(x).

Then,

L(φ(x)− yp(x)) = L(φ(x))− L(yp(x)) = r(x)− r(x) = 0.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Theorem

Let yp(x) be any solution of

y ′′ + p(x)y ′ + q(x)y = r(x)

and y1(x), y2(x) be a basis of the solution space of thecorresponding homogeneous DE.Then the set of solutions of the non-homogeneous DE is

{c1y1(x) + c2y2(x) + yp(x) | c1, c2 ∈ R}.

Proof: Let φ(x) be any solution of

L(y) = y ′′ + p(x)y ′ + q(x)y = r(x).

Then,

L(φ(x)− yp(x)) = L(φ(x))− L(yp(x)) = r(x)− r(x) = 0.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Theorem

Let yp(x) be any solution of

y ′′ + p(x)y ′ + q(x)y = r(x)

and y1(x), y2(x) be a basis of the solution space of thecorresponding homogeneous DE.Then the set of solutions of the non-homogeneous DE is

{c1y1(x) + c2y2(x) + yp(x) | c1, c2 ∈ R}.

Proof:

Let φ(x) be any solution of

L(y) = y ′′ + p(x)y ′ + q(x)y = r(x).

Then,

L(φ(x)− yp(x)) = L(φ(x))− L(yp(x)) = r(x)− r(x) = 0.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Theorem

Let yp(x) be any solution of

y ′′ + p(x)y ′ + q(x)y = r(x)

and y1(x), y2(x) be a basis of the solution space of thecorresponding homogeneous DE.Then the set of solutions of the non-homogeneous DE is

{c1y1(x) + c2y2(x) + yp(x) | c1, c2 ∈ R}.

Proof: Let φ(x) be any solution of

L(y) = y ′′ + p(x)y ′ + q(x)y = r(x).

Then,

L(φ(x)− yp(x)) = L(φ(x))− L(yp(x)) = r(x)− r(x) = 0.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Theorem

Let yp(x) be any solution of

y ′′ + p(x)y ′ + q(x)y = r(x)

and y1(x), y2(x) be a basis of the solution space of thecorresponding homogeneous DE.Then the set of solutions of the non-homogeneous DE is

{c1y1(x) + c2y2(x) + yp(x) | c1, c2 ∈ R}.

Proof: Let φ(x) be any solution of

L(y) = y ′′ + p(x)y ′ + q(x)y = r(x).

Then,

L(φ(x)− yp(x))

= L(φ(x))− L(yp(x)) = r(x)− r(x) = 0.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Theorem

Let yp(x) be any solution of

y ′′ + p(x)y ′ + q(x)y = r(x)

and y1(x), y2(x) be a basis of the solution space of thecorresponding homogeneous DE.Then the set of solutions of the non-homogeneous DE is

{c1y1(x) + c2y2(x) + yp(x) | c1, c2 ∈ R}.

Proof: Let φ(x) be any solution of

L(y) = y ′′ + p(x)y ′ + q(x)y = r(x).

Then,

L(φ(x)− yp(x)) = L(φ(x))− L(yp(x))

= r(x)− r(x) = 0.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Theorem

Let yp(x) be any solution of

y ′′ + p(x)y ′ + q(x)y = r(x)

and y1(x), y2(x) be a basis of the solution space of thecorresponding homogeneous DE.Then the set of solutions of the non-homogeneous DE is

{c1y1(x) + c2y2(x) + yp(x) | c1, c2 ∈ R}.

Proof: Let φ(x) be any solution of

L(y) = y ′′ + p(x)y ′ + q(x)y = r(x).

Then,

L(φ(x)− yp(x)) = L(φ(x))− L(yp(x)) = r(x)− r(x) = 0.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Hence, φ(x)− yp(x) is a solution of the homogeneous DE.

Thus,

φ(x)− yp(x) = c1y1(x) + c2y2(x),

for c1, c2 ∈ R. Hence,

φ(x) = c1y1(x) + c2y2(x) + yp(x).

Summary: In order to find the general solution of anon-homogeneous DE, we need to

get one particular solution of the non-homogeneous DE

get the general solution of the corresponding homogeneousDE.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Hence, φ(x)− yp(x) is a solution of the homogeneous DE. Thus,

φ(x)− yp(x) = c1y1(x) + c2y2(x),

for c1, c2 ∈ R.

Hence,

φ(x) = c1y1(x) + c2y2(x) + yp(x).

Summary: In order to find the general solution of anon-homogeneous DE, we need to

get one particular solution of the non-homogeneous DE

get the general solution of the corresponding homogeneousDE.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Hence, φ(x)− yp(x) is a solution of the homogeneous DE. Thus,

φ(x)− yp(x) = c1y1(x) + c2y2(x),

for c1, c2 ∈ R. Hence,

φ(x) = c1y1(x) + c2y2(x) + yp(x).

Summary: In order to find the general solution of anon-homogeneous DE, we need to

get one particular solution of the non-homogeneous DE

get the general solution of the corresponding homogeneousDE.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Hence, φ(x)− yp(x) is a solution of the homogeneous DE. Thus,

φ(x)− yp(x) = c1y1(x) + c2y2(x),

for c1, c2 ∈ R. Hence,

φ(x) = c1y1(x) + c2y2(x) + yp(x).

Summary:

In order to find the general solution of anon-homogeneous DE, we need to

get one particular solution of the non-homogeneous DE

get the general solution of the corresponding homogeneousDE.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Hence, φ(x)− yp(x) is a solution of the homogeneous DE. Thus,

φ(x)− yp(x) = c1y1(x) + c2y2(x),

for c1, c2 ∈ R. Hence,

φ(x) = c1y1(x) + c2y2(x) + yp(x).

Summary: In order to find the general solution of anon-homogeneous DE, we need to

get one particular solution of the non-homogeneous DE

get the general solution of the corresponding homogeneousDE.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Hence, φ(x)− yp(x) is a solution of the homogeneous DE. Thus,

φ(x)− yp(x) = c1y1(x) + c2y2(x),

for c1, c2 ∈ R. Hence,

φ(x) = c1y1(x) + c2y2(x) + yp(x).

Summary: In order to find the general solution of anon-homogeneous DE, we need to

get one particular solution of the non-homogeneous DE

get the general solution of the corresponding homogeneousDE.

Santanu Dey Lecture 7

Non-homogeneous Second Order Linear ODE’s

Hence, φ(x)− yp(x) is a solution of the homogeneous DE. Thus,

φ(x)− yp(x) = c1y1(x) + c2y2(x),

for c1, c2 ∈ R. Hence,

φ(x) = c1y1(x) + c2y2(x) + yp(x).

Summary: In order to find the general solution of anon-homogeneous DE, we need to

get one particular solution of the non-homogeneous DE

get the general solution of the corresponding homogeneousDE.

Santanu Dey Lecture 7