Download - DE Lecture 3
-
7/30/2019 DE Lecture 3
1/2
7
Exercises
I. Supply information to each DE in the indicated title:type (ode or pde), order(1st , 2nd, ), degree (1st,2nd, ), linearity (linear or nonlinear), dependent & independent variables (write the variable/spresent in the equation).
DE Type Order Degree LinearityDependentVariable(s)
IndependentVariable
1 y = 3x + e2x
2 tzdt
dz
dt
zd
2
3
3
4
3 zy
zx
x
z
2
2
2
2
4 1232
2
2
dx
dy
dx
ydey
5 0232
r
v
t
v
rt
v
6 wzdz
dwz
dz
wd54
3
3
sin
7 y 3xy + 4x2y y = ex + 2
8 032
2
3
32 ty
dt
ydt
dt
ydt cossin
9 122
2
xdx
dy
dx
ydey
10 tdtdyy
dtdy 35
II. Verify whether each given function is an explicit solution to the DE.1. y = x2 + 3x 1; x2y xy + y = x2 12. y = 4sin x 2x + 2; y + y + y = 4cosx 2x3. y = e3x 3cosx + 4x; y 3y + 3y 3y = 4 12x4. y1 = 2ex , y2 = 4ex , y3 = 31 ex; y + y = 05. y1 =
x
1, y2 =
x
2for all x 0, y3 =
2
1
x
for all x 2; y + y2 = 0
6. x1(t) = 2sin 2t, x2(t) = cos 2t, x3(t) = 3 cos 2t; 042
2
xdt
xd
7. y1 = ex x 1, y2 = cos x + sin x + ex x 1; y y + y y = x8.
214
x
xy
ln ,
226
xy ; x4y + 4x3y + 4x2y + 4 = 0, (x > 0)
-
7/30/2019 DE Lecture 3
2/2
8
III. Determine whether the given function is the implicit solution of the given DE.1. x2y + 6 = 0, x2y + 4xy + 2y = 02. yex + xy = 1, y(x + ex) + 2y(1 + ex) + yex = 03. t2z 4 ln t = 0, 045
2
22 z
dt
dzt
dt
zdt
4. 3x2 + 5y2 +3x 10y 15 = 0, 3yy + y(y 1) =05. rsin + cos 2r+ 5 = 0, 0122
2
23
sinsinsincossin rd
dr
d
rd
d
rd
IV. Show that the given function is the general solution of the DE.1. y = Cex, y + y = 02.
kxy
1, y + y2 = 0
3. (y A)2 =Ax, 4x(y)2 + 2xy y = 04. y = C1 sin 2x + C2 cos 2x, y + 4y = 05. y =Aet + Be2t + t, 3223
2
2
xydt
dy
dt
yd
6. y = C1 sin x + C2x , (1 x cot x)y xy + y = 07. x = C1e2v + C2e2v + C3e4v , 01644
2
2
3
3
ydv
dx
dv
xd
dv
xd
8. y =Aex+ Bxex +Cex +2x2ex, y y y + y = 8ex