Page 24 - Pra U STPM 2022 Penggal 2 - Mathematics

Mathematics Semester 2 STPM Chapter 4 Differential Equations

Example 6

dy 1 + y 2
Find the general solution of the differential equation = .
dx 2y
Hence, find the particular solution if y = 0 when x = 0.
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Solution: From dy = 1 + y 2 ,
dx 2y
2y
1 + y 2 dy = dx
2y
∫ 1 + y 2 ∫
= dx
2
ln |1 + y | = x + C where C is an arbitrary constant
1 + y = e x + C
2
x
2
y = Ae – 1 where A = e C
Ae –
Thus, the general solution of the differential equation is y = ± 1 .
x
When x = 0, y = 0, C = 0 and A = 1.
x
e –
Therefore, the particular solution of the differential equation is y = ± 1 .
The particular solution can also be obtained by using definite integration with the initial condition as the lower
limit and x and y as the upper limit.

Example 7

2y
4 The gradient function of a curve at any point (x, y) is given by the equation dy = 1 – x 2 . Find the equation
dx
2
of the curve if the curve passes through the point 1 1 , 1 .
2
Solution: The equation of the gradient of the curve can be written as
dy 2 dx
y = 1 – x 2

2 dx
∫
∫ dy = 1 – x 2
y

ln | y | = ln 1 + x  + C
1 – x
When x = 1 and y = 1, C = – ln 3
2

i.e. ln | y | = ln 1 + x  – ln 3
1 – x
2
Thus, the equation of the curve that passes through the point 1 1 , 1 is
2
1 + x
y = .
3(1 – x)

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04 STPM Math(T) T2.indd 132 28/01/2022 5:44 PM