Download Math 285 D1 Review Problems – Solutions 1. xy ′ − 3y = x2y2, y(1

Math 285 D1
Review Problems – Solutions
1. xy ′ − 3y = x2 y 2 , y(1) = 1.
Assume y ̸= 0 and rewrite as y ′ − x3 y = xy 2 . This is
a Bernoulli equation. Put v = y −1 , so v ′ = −y −2 y ′ and
y ′ = −y 2 v ′ . Thus −y 2 v ′ − x3 y = xy 2 and v ′ + x3 v =
−x.
∫ 3dx
This is first order linear: the integrating factor is e x =
e3 ln x = x3 . Hence (vx3 )′ = −x4 and vx3 = − 51 x5 + c.
Thus v = − 51 x2 + xc3 and hence
1
x3
=
,
c
x2
x5
−
c
−
3
x
5
5
and y = 0, is the general solution. Now find the solution
such that y(1) = 1. This gives 1 = c−1 1 , so c = 65 and
1
y= =
v
y=
5x3
6−x5
5
is the required solution.
2. xyy ′ = x2 + y 2 .
2
+y 2
Rewrite the equation as y ′ = x xy
, which is homoy
geneous. Put v = x , so that y = vx and y ′ = v ′ x + v.
2
2 2
x
1+v 2
1+v 2
1
′
Hence v∫′ x + v =∫ x +v
=
,
so
v
x
=
2
x v
v
v − v = v.
1 2
Hence vdv = dx
Thus
x , so 2 v = ln x + constant.
√
2
v = 2 ln x + c with c a constant, and y = ±x 2 ln x + c.
3. yy ′ = x2 + y 2 .
Put v = x2 + y 2 ; then v ′ = 2x + 2yy ′ . Hence yy ′ =
1 ′
1 ′
2 (v − 2x). Then we get 2 (v − 2x) = v, which yields
v ′ − 2v =
∫ 2x. This is first order linear with integrating
factor e −2dx = ∫e−2x . Thus we get (ve−2x )′ = 2xe−2x .
Hence ve−2x = 2 xe−2x dx. Therefore
∫
∫
1 −2x
1
2x
−2x
2x
v = 2e
xe dx = 2e {− xe
− − e−2x dx} =
2
2
1
∫
1
1
e−2x dx = −x+e2x (− e−2x +c) = −x− +ce2x
2
2
√
Hence y = ± ce2x − x − 12 − x2 .
−x+e
2x
2
3y−y
4. y ′ = 2xy−3x+2y
.
Rewrite this as (y 2 − 3y)dx + (2xy − 3x + 2y)dy = 0.
∂
∂
Now ∂y
(y 2 −3y) = 2y −3 and ∂x
(2xy −3x+2y) = 2y −3,
so the equation is exact.
Find F (x, y) such that Fx = y 2 − 3y and Fy = 2xy −
3x + 2y. From the first equation F = xy 2 − 3xy + g(y),
so that Fy = 2xy − 3x + g ′ (y) = 2xy − 3x + 2y. Hence
g ′ (y) = 2y and g(y) = y 2 . Thus F = xy 2 − 3xy + y 2 . The
general solution is therefore xy 2 − 3xy + y 2 = c.
5. y ′ = xy − y 2 + 1.
This is a Riccati equation. Note that y1 = x is a
′
solution. Put y = y1 + v1 = x + v1 . Then y ′ = 1 − vv2 .
Hence
v′
1
1
1 − 2 = x(x + ) − (x + )2 + 1,
v
v
v
v′
x
1
which leads to − v2 = − v − v2 . Hence v ′ −xv
= 1, which is
∫
2
−xdx
first order linear with integrating factor ∫e
= e−x /2 .
2
2
2
2
Thus (ve−x /2 )′ = e−x /2 , so v = ex /2 e−x /2 dx. The
general solution is therefore
y =x+
e
1 2
2x
∫
1
e− 2 x dx
1 2
.
6. xy ′′ = 1 + (y ′ )2 .
Put p ∫= y ′ , so that y ′′ = p′ and xp′ = 1 + p2 . Thus
∫ dp
dx
−1
p = ln x + c1 . Thus p =
1+p2 =
x and hence tan
2
tan(ln x + c1 ). The general solution is
∫
y = tan(ln x + c1 )dx.
7. yy ′′ = (y ′ )2 .
dp
dp
Put p = y ′ ; then y ′′ = p dy
and yp dy
= p2 . If p = 0,
dp
then y is constant. Let p ̸= 0, so that dy
= yp and
∫ dp
∫ dy
=
. Thus ln p = ln y + constant and p = c1 y.
p
∫ dyy ∫
Hence y = c1 dx and we get ln y = c1 x + constant.
Hence y = c2 ec1 x , which includes the case where y is
constant.
8. (2xy − y 3 + y 2 )dx − (x2 + xy 2 − y 3 )dy = 0.
Look for an integrating factor of the form µ = µ(y).
Multiply the equation by µ to get
µ · (2xy − y 3 + y 2 )dx + µ · (−x2 − xy 2 + y 3 )dy = 0.
The condition for exactness is
µ′ · (2xy − y 3 + y 2 ) + µ · (2x − 3y 2 + 2y) = µ · (−2x − y 2 ).
Hence µ′ · (2xy − y 3 + y 2 ) = µ · (−4x − 2y + 2y 2 ) and
µ′ · y(2x − y 2 + y)
= −2µ∫· (2x − y 2 + y). Thus we have
∫
2dy
µ′ ·y = −2µ and dµ
µ =−
y . This gives ln µ = −2 ln y
and so µ = y −2 is an integrating factor. Multiply the
original equation by y −2 to get the exact equation
2x
x2
( − y + 1)dx + (− 2 − x + y)dy = 0.
y
y
Now find F (x, y) such that Fx =
2
− xy2 − x + y. Thus F =
2
x
y
3
2x
y
− y + 1 and Fy =
− xy + x + g(y) and Fy =
− xy2 − x + g ′ (y) = − xy2 − x + y. Hence g ′ = y and g =
2
Then F =
2
x
y
2
x2
y
− xy + x +
− xy + x +
2
y
2
y2
2.
y2
2.
The general solution is
= c.
9. y ′′ + 3y ′ + 2y = x + e−x .
The homogeneous equation is y ′′ + 3y ′ + 2y = 0, with
characteristic polynomial f (r) = r2 + 3r + 2 = 0; its
roots are −1, −2. Thus yh = c1 e−x + c2 e−2x . Now find a
particular solution of y ′′ + 3y ′ + 2y = x. Try y = ax + b;
then y ′ = a, y ′′ = 0 Substitute to get a = 21 , b = − 43 .
Thus 12 x − 34 is a particular solution.
Next a particular solution of y ′′ + 3y ′ + 2y = e−x is
xe−x
−x
2
f ′ (−1) = xe , since −1 is a root of r + 3r + 2 with
multiplicity 1. Hence a particular solution of y ′′ + 3y ′ +
2y = x + e−x is yp = 21 x − 34 + xe−x . The general solution
is
1
3
y = c1 e−x + c2 e−2x + x − + xe−x .
2
4
10. y ′′ − 4y ′ + 5y = e2x sin x.
The characteristic polynomial is f (r) = r2 − 4r + 5,
which has roots 2 ± i. Then yh = (c1 cos x + c2 sin x)e2x .
To find a particular solution write down the cosine equation v ′′ − 4v ′ + 5v = e2x cos x. Add i times the sine
equation to the cosine equation and write z = v + iy.
This gives
z ′′ − 4z ′ + 5z = e(2+i)x .
2+ix
A particular solution of this equation is zp = fxe′ (2+i) , since
2+i is a root of f with multiplicity 1. Hence a particular
4
solution of the given equation is yp = Im(zp ). Thus
xe2+ix
cos x + i sin x
1
yp = Im(
) = xe2x Im(
) = − xe2x cos x.
2i
2i
2
The general solution is y = (c1 cos x+c2 sin x)e2x − 12 xe2x cos x.
11. x2 y ′′ + 2xy ′ − 2y = sin x.
First the homogeneous Euler equation is x2 y ′′ +2xy ′ −
2y = 0. Look for solutions y = xr ; then y ′ = rxr−1 ,
y ′′ = r(r − 1)xr−2 . Substitute in the equation to get
r(r − 1) + 2r − 2 = 0, i.e., r2 + r − 2 = 0. This has roots
1, −2, so two linearly independent solutions are y1 = x
and y2 = x−2 . Hence yh = c1 x + xc22 .
Now use variation of parameters to find a particular
solution y = v1 y1 + v2 y2 . Note that W r(y1 , y2 ) = −3x−2 .
Then we have
∫
∫ −2
1
x sin x
1
sin
xdx
=
−
v1 = −
dx
=
cos x.
−3x−2
3
3
∫
x sin x
1
v2 =
sin3 xdx.
dx = −
−2
−3x
3
By
successive integration by parts we find that
∫ three
3
sin xdx = −x3 cos x + 3x2 sin x + 6x cos x − 6 sin x.
Therefore
1
v2 = x3 cos x − x2 sin x − 2x cos x + 2 sin x.
3
A particular solution of the equation is
Also
∫
1
1
yp = (− cos x)x+( x3 cos x−x2 sin x−2x cos x+2 sin x)x−2 ,
3
3
5
x
2 sin x
which reduces to yp = − sin x− 2 cos
x + x2 . The general
solution is
c2
2 cos x 2 sin x
y = c1 x + 2 − sin x −
+
.
x
x
x2
6