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Mathematics Semester 2 STPM Chapter 4 Differential Equations
Example 12
dy x + y + 5 dv 2(v + 1)
If v = x + y, show that the equation = x + y – 3 can be reduced to = v – 3 , and hence solve
the equation. dx dx
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dy x + y + 5
Solution: =
dx x + y – 3
Substituting v = x + y or y = v – x,
d v + 5
(v – x) =
dx v – 3
dv – 1 = v + 5
dx v – 3
dv = v + 5 + 1
dx v – 3
= 2(v + 1)
v – 3
∫
∫ v – 3 dv = 2 dx
v + 1
∫
4
∫1 v + 1 2 dv = 2 dx
1 –
v – 4 ln | v + 1 | = 2x + C
x + y – 4 ln | x + y + 1 | = 2x + C
1
ln | x + y + 1 | = 1 (y – x) + D D = – C
4 4
1
x + y + 1 = Ae —(y – x) A = e D
4
1
x + y + 1 – Ae —(y – x) = 0
4
4
Example 13
3
By writing u = 1 , reduce the equation 2 dy – y = y to du + u = –1.
y 2 dx x dx x
Hence solve the differential equation.
Solution: Differentiating u = 1 with respect to x
y 2
du d 2 dy
–2
= y = –
dx dy y 3 dx
dy y
3
Dividing the equation 2 – = y with y .
3
dx x
2 dy – 1 = 1
y 3 dx xy 2
du u
\ – – = 1
dx x
du u
\ + = –1
dx x
1 dx
∫
ln x
The integrating factor is e x = e = x.
138
04 STPM Math(T) T2.indd 138 28/01/2022 5:44 PM

