TWK2A

Solutions by substitutions
Solutions
1. Letting x = vy we have
y(vdy + ydv) 2(vy + y)dy = 0
ydv (v + 2)dy = 0
dv dy
) = 0
v+2 y
) ln jv + 2j ln jyj = c
x
) ln + 2 ln jyj = c
y
) x + 2y = cy 2

2. Letting y = ux we have
(x + 3ux)dx (3x + ux)(udx + xdu) = 0
) (u2 1)dx + x(u + 3)du = 0
dx u+3
) + du = 0
x (u 1)(u + 1)
) ln jxj + 2 ln ju 1j ln ju + 1j = c
x(u 1)2
) = c
u+1
y 2 y
)x 1 = c +1
x x
) (y x)2 = c(y + x)

3. Letting y = ux we have
(x + uxeu )dx xeu (udx + xdu) = 0
) dx xeu du = 0
dx
) eu du = 0
x
) ln jxj eu = c
) ln jxj ey=x = c:
4. From y 0 y = ex y 2 and w = y 1
we obtain
dw
+w = ex :
dx
An integrating factor is ex so that
1 2x 1 x
ex w = e +c)y 1
= e + ce x :
2 2
3
5. From y 0 + y = y 1=2
and w = y 2 we obtain
dw 3 3
+ w= :
dx 2 2
An integrating factor is e3x=2 so that
3x
3x 3 3x
e 2 w = 3 2 +c ) y 2 = 1 + ce 2 :

If y(0) = 4 then c = 7 and

3 3x
y 2 = 1 + 7e 2 :

6. With u = y x + 5 we have
du dy
= 1:
dx dx
Hence,
du
+ 1 = 1 + eu ) e u
du = dx:
dx
u
Thus, e = x + c and so
(y x+5)
e = x + c ) y (x) = x 5 ln ( x c) :

Note that for this solution the interval of de…nition must be such that x
c > 0 ) x < c: In other words, the interval of de…nition of the solution
here is in‡uenced by the value of the integration constant.
7. The substitutions y = y1 + u and
dy dy1 du
= +
dx dx dx
lead to
dy1 du
+ = P + Q(y1 + u) + R(y1 + u)2
dx dx
= P + Qy1 + Ry12 + Qu + 2y1 Ru + Ru2

or
du
(Q + 2y1 R)u = Ru2 :
dx
This is a Bernoulli equation with n = 2 which can be reduced to the linear
equation
dw
+ (Q + 2y1 R)w = R
dx
by the substitution w = u 1 . For the given DE, identify P (x) = 4=x2 ; Q(x) =
1=x; and R(x) = 1: Then dw dx
+ 1
x
+ x4 w = 1: An integrating factor is
1
x3 so that x3 w = 41 x4 + c or u = 1
4
x + cx3 . Thus,
1
2 1
y (x) = + x + cx3 :
x 4

8.
dy p
x y= x2 + y 2 (1)
dx
Rearranging:
p
y+ x2 + y 2 dx xdy = 0
For p
M (x; y) y+ x2 + y 2 and N (x; y) x
we have
p p
M (tx; ty) = ty + t2 x2 + t2 y 2 = t y + x2 + y 2 = tM (x; y)
N (tx; ty) = t( x) = tN (x; y):
 Thus, (1) is homogeneous and we use the substitution
y
u= (2)
x
which implies
y = ux
dy : (3)
dx
= y + x du
dx

Substitute (3) into (1):

du p
x u+x ux = x 2 + x 2 u2
dx
du p
x = 1 + u2
dx
Separate variables and integrate:
Z Z
du dx
p = + c1 (4)
1+u 2 x
p
) ln u + 1 + u2 = ln jxj + c1
p
u + 1 + u2
) ln = c1
x

The integral on the LHS of (4) may be obtained using the substitution
u = tan . Now use (2):
q
y 2
x
+ 1 + xy 2
ln = c1
x
p
y + x2 + y 2
) ln = c1
x
p
) y + x2 + y 2 = c2 x

9.
dx
y + x(ln x ln y 1) = 0 (5)
dy
This equation is neither homogeneous nor is it a Bernoulli equation. How-
ever, since (5) can be written in the form
dx x
y + ln 1 =0 (6)
dy y
it is natural to try the substitution
x
u= (7)
y
Then we have
x = uy
dx : (8)
dy
= u + y du
dx

Substitute (8) into (6):

du
y u+y + uy (ln u 1) = 0
dy
du
)y + u ln u = 0
dy
Separate variables and integrate:
Z 1 Z
u dy
du + = c1
ln u y
) ln jln jujj + ln jyj = c1
) ln jy ln jujj = c1
) y ln juj = c
Substitute (7):
x
y ln =c
y
Initial value: x = 1 ; y = e :
1
c = e ln = e
e
The solution is
x
y ln = e
y
or e
x = ye y :
10.
dy
x+y+x =0 (9)
dx
Since
M (tx; ty) = tx + ty = t(x + y) = tM (x; y)
and
N (tx; ty) = tx = tN (x; y)
both functions are homogeneous functions of degree 1. We therefore let
dy du
y = ux ; =u+x (10)
dx dx
Substitute (10) in (9):
du
x + ux + x u + x = 0
dx
du
) 1 + 2u + x = 0
dx
Separate variables and integrate:
Z Z
dx du
+ = c1
x 1 + 2u
1
) ln jxj + ln j1 + 2uj = c1
2
) 2 ln jxj + ln j1 + 2uj = c2
) ln jx2 (1 + 2u)j = c2
) x2 (1 + 2u) = c
y
Substitute u = x
:
y
x2 1 + 2 = c
x
1 c
) y (x) = x :
2 x
11.
dy
3 1 + t2 = 2ty(y 3 1)
dt
dy 2t 2t
) + 2
y = y4 (11)
dt 3(1 + t ) (1 + t2 )
 This is a Bernoulli equation with n = 4. We therefore let
1
u = u1 4
= (12)
y3
Then follows that
du 3 dy 3u dy
= =
dt y 4 dt y dt
and therefore
dy y du
= (13)
dt 3u dt
Substitute (12) and (13) in (11):
y du 2t 2t y
+ y =
3u dt 3(1 + t2 ) 3(1 + t2 ) u
1 du 2t 1
) + 2
1 = 0
u dt (1 + t ) u
Separate variables and integrate:
Z Z
du 2tdt
+ = c1
1 u 1 + t2
) ln j1 uj + ln(1 + t2 ) = c1
1 + t2
) ln = c1
1 u
1 + t2
) = c2
1 u
)1 u = c(1 + t2 )
1
Substitute u = y3
:

1
1 = c(1 + t2 )
y3
1
) y3 =
1 c(1 + t2 )

12. We have the equation

(y 2 + xy)dx x2 dy = 0
 where we identify M (x; y) y 2 + xy and N (x; y) x2 . Since

M (tx; ty) = t2 y 2 + txty = t2 (y 2 + xy) = t2 M (x; y)

is homogeneous of order 2, and

N (tx; ty) = t2 x2 = t2 ( x2 ) = t2 N (x; y)

is homogeneous of order 2, we have a homogeneous di¤erential equation.

Thus we use the substitution
dy du
y(x) = u(x)x; = u(x) + x ; dy = u dx + x du
dx dx
so that the equation
(y 2 + xy)dx x2 dy = 0
becomes
(u2 x2 + ux2 )dx x2 (u dx + x du) = 0
and grouping di¤erentials we …nd

u2 x2 dx = x3 du:

Integration yields Z Z
dx du
= +c
x u2
1 x
) ln jxj = +c= +c
u y
since u = y=x. Thus the solution is given by
x
+ ln jxj = c:
y

13. Rewriting the equation as

p
(y + x2 y 2 )dx xdy = 0
p
with M (x; y) y + x2 y 2 and N (x; y) x we …nd that
p p
M (tx; ty) = ty + t2 x2 t2 y 2 = t(y + x2 y 2 ) = tM (x; y); t>0
 is of homogeneous of order 1 for t 2 [0; 1); and

N (tx; ty) = tx = tN (x; y)

is also homogeneous of order 1, so that the equation is a homogeneous dif-

ferential equation. Using
dy du
y(x) = u(x)x; = u(x) + x ; dy = u dx + x du
dx dx
we …nd p
ux + x2 u2 x2 dx x(u dx + x du) = 0
p
) x 1 u2 dx = x2 du
since x > 0. Next we separate and integrate
Z Z
du dx
p = +c
1 u 2 x

) arcsin(u) = ln jxj + c
y
) arcsin = ln jxj + c:
x
14. Using the substitution
du dy
u(x; y) x + y; =1+
dx dx
we obtain
du 1 u 1
1= = 1
dx u u
which is trivially separated to obtain
1 2
u = x+c
2
) (x + y)2 = 2x + 2c:

15.
p p
ydx + (x + xy) dy = 0 ) M (x; y) = y and N (x; y) = x + xy
 M (tx; ty) = ty = t ( y) = tM (x; y)
p p p
N (tx; ty) = tx + txty = tx + t2 xy = tx + t xy = tN (x; y)
) both M and N are homogeneous functions of degree 1.

The substitution y = ux gives

p
uxdx + x + x u (udx + xdu) = 0
p
) x2 + x2 u du + xu3=2 dx = 0
1 dx
) u 3=2 + du + =0
u x
) 2u 1=2 + ln juj + ln jxj = c
r
y x
) ln + ln jxj = 2 +c
x y
) y (ln jyj c)2 = 4x

16.
dy 1
x +y = 2
dx y
dy y y 2
) + = ) Bernoulli DE with n = 2
dx x x
1 ( 2) 3 1=3 dy u 2=3 du
So u = y =y )y=u ) =
dx 3 dx
2=3
u du u1=3 u 2=3 du 3 3
Hence, + = ) + u=
3 dx x x dx x x
R 3
dx
which is a linear …rst-order DE with integrating factor e x = x3 :
Z
3
) x u= 3x2 dx = x3 + c

) x3 y 3 = x3 + c
) y 3 = 1 + cx 3 :

17.
dy
x2 2xy = 3y 4
dx
 dy 2y 3y 4
) = 2 ) Bernoulli DE with n = 4
dx x x
dy u 4=3 du
So u = y 1 (4) = y 3 ) y = u 1=3 ) =
dx 3 dx
4=3 1=3
u du 2u 3u 4=3 du 6 9
Hence, = ) + u=
3 dx x x2 dx x x2
R 6
dx
which is a linear …rst-order DE with integrating factor e x = x6 :
Z
6 9x3
) x u= 9 x4 dx = +c
5
9
) u= + cx 6
5x
3 9
) y = + cx 6 :
5x

3
1 1 9
Initial value: y (1) = ) = + c(1)
2 2 5(1)
9
) 8= +c
5
49
) c=
5
9 1 49 6
) y 3= x + x :
5 5