Applied maths/AL/Tutorial/De-2/sol/p.1
PLK Vicwood K.T.Chong Sixth Form College
Applied Mathematics (AL)
Solution to Tutorial
Topic: First order linear equations
The reduction of second order equations to first order equations
1. (a)
Code: De-2
dy
+ y = xex + 1.
dx
 1dx = x + C
where C0 is a arbitrary constant.
0
An integrating factor is ex.
Multiplying both sides by ex,
dy
ex
+ exy = xe2x + ex.
dx
d (e x y )
= xe2x + ex.
dx
exy =
 ( xe
2x
 e x )dx

1
xde2 x + ex
2
1
1
e 2 x dx + ex
= xe2 x 
2
2
1
1
= xe2 x  e2x + ex + C
2
4
1 x 1 x
= xe  e + 1 + Cex.
2
4
=

y
(b)
where C is a arbitrary constant.
dy
 y = 2ex.
dx
 (1)dx = x + C
0
where C0 is a arbitrary constant.
An integrating factor is ex.
Multiplying both sides by ex,
dy
ex
 exy = 2.
dx
d (e  x y)
= 2.
dx
exy =
y

2dx
= 2x + C.
= 2xex + Cex
where C is a arbitrary constant.
Applied maths/AL/Tutorial/De-2/sol/p.2
(c)
2
dy
 2 xy  2 xe x .
dx

2 xdx = x2 + C0
where C0 is a arbitrary constant.
2
An integrating factor is e x .
2
Multiplying both sides by e x ,
2 dy
2
ex
 2 xe x y = 2x.
dx
2
d (e x y )
= 2x.
dx
2
ex y =

2xdx
= x2 + C.
y
(d)
= x2 e  x  Ce  x
2
2
where C is a arbitrary constant.
dy
+ 4x1y = x4.
dx
 (4 x
1
) dx = 4ln|x| + C0 = ln|x4| + C0
where C0 is a arbitrary constant.
An integrating factor is x4.
Multiplying both sides by x4,
dy
x4
+ 4x3y = x8.
dx
d ( x 4 y)
= x8.
dx
x9
x4y =
+C
9
1
C
y = x5 
.
9
x4
where C is a arbitrary constant.
dy
+ 2xy  y3 = 0 ; x > 0.
dx
Clearly, y = 0 is a trivial solution.
To obtain other solutions, let u = y1 3 = y2.
du
dy
= 2y3
dx
dx
2. (a) x2
= 2y3(
y3
2

x
2 4
 u.
= 
x2 x
2
y)
x
Applied maths/AL/Tutorial/De-2/sol/p.3
du 4
2
 u
.
dx x
x2
(*)
 ( x )dx = 4ln|x| + C
4
0
= ln |x4| + C0
An integrating factor is x4.
Multiplying both sides of (*) by x4,
du
x4
 4x5 = 2x6.
dx
where C0 is a arbitrary constant.
d ( x 4u )
= 2x6.
dx
x4u =
 (2 x
6
=
)dx
2 5
x +C
5
where C is a arbitrary constant.
Therefore, either y = 0 or y 2 =
(b)
5x
2  C1 x 5
where C1 = 5C.
dy
= ry  ky2 where r and k are positive constants.
dx
Clearly, y = 0 is a trivial solution.
To obtain other solutions, let u = y12 = y1.
du
dy
= y2
dx
dx
= y2(ry  ky2)
= ru + k.
du
+ ru = k.
(*)
dx

rdx
= rx + C0
where C0 is a arbitrary constant.
An integrating factor is erx.
Multiplying both sides of (*) by erx,
du
erx
+ rerxu = kerx.
dx
d (e rx u )
= kerx.
dx
erxu =

kerx dx
=
k rx
e +C
r
Therefore, either y = 0 or y =
where C is a arbitrary constant.
r
k  C1e  rx
where C1 = rC.
Applied maths/AL/Tutorial/De-2/sol/p.4
(c)
1
dy 3
 y  x4 y 3 .
dx x
Clearly, y = 0 is a trivial solution.
1 1
2
To obtain other solutions, let u = y 3  y 3 .
du
2  1 dy
= y 3
dx
3
dx
2  13 4 13 3
= y ( x y  y)
3
x
2 4 2
= x  u.
3
x
du 2
2
 u  x4
(*)
dx x
3
2
( )dx = 2ln|x| + C0 = ln|x2| + C0
where C0 is a arbitrary constant.
x
An integrating factor is x2.
Multiplying both sides of (*) by x2,
du 2
2

u  x2.
x2
dx x 3
3

d ( x 2u) 2 2
 x.
dx
3
2
x 2 dx
x2u =
3
2
= x3 + C
9

where C is a arbitrary constant.
2
Therefore, either y = 0 or
(d)
y3 
2 5
x  Cx 2 .
9
dy
= (cost + T)y  y3 where  and T are constants and y = y0  0 when t = 0.
dt
Clearly, the trivial solution y = 0 does not satisfy the condition y(0) = y0  0.
To obtain other solution, let u = y13 = y2.
du
dy
= 2y3
dt
dt
3

= 2y ((cost + T)y  y3)
= 2(cost + T)u +2.
du
+ 2(cost + T)u = 2.
dt
 2( cos t  Tt )dt = 2(sin t + T t) + C
An integrating factor is e2(sin t + T t).
0
(*)
where C0 is a arbitrary constant.
Applied maths/AL/Tutorial/De-2/sol/p.5
Multiplying both sides of (*) by e2(sin t + T t),
du
e2(sin t + T t)
+ 2(cost + T)e2(sin t + T t)u = 2e2(sin t + T t).
dt
d
(ue2(sin t + T t)) = 2e2(sin t + T t).
dt
u(t)e2(sin t + T t ) u(0)e2(sin (0) + T(0)) =
Therefore, y2 =
t 2( sin s  Ts )
e
ds .
0

e 2( sin t  Tt )
t 2( sin s  Ts )
y 0 2 
e
ds
0

.
y12
2x  1
(
) y1  x  2
x
x
x2
2x  1
=
(
)x  x  2
x
x
= 1
dy1
=
.
dx
y1(x) = x is a particular solution of the differential equation.
1
(b) To find other solutions, put y = x + .
u
1 du
1
1 2 2x  1
1
( x  ) + x + 2.
1
= (x  ) 
x
u
x
u
u 2 dx
1 du 2
1
2x  1

 

.
2 dx
2
u xu
xu
u
du 1
1
 u .
(*)
dx x
x
1
( )dx = ln|x| + C0 = ln|x1| + C0
where C0 is a arbitrary constant.
x
An integrating factor is x1.
Multiplying both sides of (*) by x1,
du 2
x1
x u = x2.
dx
3. (a)

d ( x 1u )
= x2.
dx
x1u = x1 + C
u = 1 + Cx.
Therefore, either y = x or
where C is a arbitrary constant.
y=x+
1
.
1  Cx
Applied maths/AL/Tutorial/De-2/sol/p.6
4.
2 cos 2 x  sin 2 x  sin 2 x
d sin x
.
 cos x 
2 cos x
dx
y = sin x is a particular solution of
dy 2 cos2 x  sin 2 x  y 2
.

dx
2 cos x
1
To find other solutions, put y = sin x + .
u
1 du
sin 2 x
1
1
cos x 
= cos x 

(sin x  ) 2 .
2 dx
2 cos x 2 cos x
u
u
1 du tan x
1



.
u
u 2 dx
2u 2 cos x
du
1
 (tan x)u   sec x .
(*)
dx
2

tan xdx
= ln |sec x| + C0
where C0 is a arbitrary constant.
An integrating factor is sec x.
Multiplying both sides of (*) by sec x,
du
1
sec x + u sec x tan x =  sec2x.
dx
2
d
1
(u sec x) =  sec2x.
dx
2
1
sec 2 xdx .
u sec x = 
2
1
=  tan x + C
2
1
u =  sin x + C cos x.
2

Therefore, either y = sec x or y = sec x +
where C is a arbitrary constant.
2
 sin x  C1 cos x
dy
.
dx
The equation becomes
du
 mu = kemx.
dx
where C1 = 2C.
5. (a) Let u =
 (m)dx =  mx + C
0
An integrating factor is emx.
Multiplying both sides of (*) by emx,
du
emx
 memxu = k.
dx
d
(ue  mx ) = k.
dx
(*)
where C0 is a arbitrary constant.
Applied maths/AL/Tutorial/De-2/sol/p.7
u emx =

kdx .
= kx + C1
dy
= u = kxemx + C1emx.
dx
y
=
 (kxe
mx
where C1 is a arbitrary constant.
 C1e mx )dx

C1 mx
e  k xdemx
m
C
k
k
e mx dx
= 1 e mx  xemx 
m
m
m
C
k
k mx
= 1 e mx  xemx 
e + C2
m
m
m2
=

(b)
d2y
2 dy
= 1.
x dx
dx
dy
Let u =
.
dx
The equation becomes
du
2
 u = 1.
dx
x
2

where C2 is a arbitrary constant.

dx =  2ln|x| + C0= ln |x2| + C0
(*)
where C0 is a arbitrary constant.
An integrating factor is x2.
Multiplying both sides of (*) by x2,
du
x2
 2x3u = x2.
dx
d
(ux  2 ) = x2.
dx
u x2
=

x  2 dx .
= x1 + C1
where C1 is a arbitrary constant.
dy
= u = x + C1x2.
dx
y
=
 ( x  C1x
=
2
)dx
x 2 C1 3

x + C2
2
3
where C2 is a arbitrary constant.
Applied maths/AL/Tutorial/De-2/sol/p.8
(c)
dy
dy d 2 y
= y2,
= y(0) = 1.
dx x  0
dx dx 2
dy
Let u =
.
dx
d2y

du du dy
du
.

u
dx dy dx
dy
dx
2
u2
du
= y2.
dy

u 2 du 
y
2
dy .
u3 y3
+ C1

3
3
dy
Since u(0) =
= y(0) = 1, C1 = 0.
dx x  0
dy
Hence
= u = y.
dx
dy
 dx .
y
ln|y| = x+ C2
y = Aex
Since y(0) = 1, A = 1.
y = ex.
for some constant C1.

for some constant C2.