Page 20 - Pra U STPM 2022 Penggal 2 - Mathematics

Mathematics Semester 2 STPM Chapter 4 Differential Equations

Solution: (a) order : 1, degree : 1
(b) order : 1, degree : 2
(c) order : 2
3
d y dy —
2
1
2
From = 1 + 2
dx 2 dx
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3
2
— 2
22 31
1 d y 2 = 1 + dy 2 4
2
dx dx
2
1
22
1 d y 2 = 1 + dy 2 3
dx
dx
Thus, the degree is 2.
Solution of differential equations
The function y = φ(x) is the solution of a differential equation if it satisfies a given differential equation.
Example 2

2
Show that y = 1 x + C is the solution of the differential equation dy = x.
2 dx
Solution: From y = 1 x + C,
2
2
dy = x
dx
2
Thus, y = 1 x + C is the solution of the differential equation dy = x.
4 2 dx

Example 3

dy
2x
Show that y = Ae + (x + 2)e is a solution of the differential equation – y = (x + 3)e .
x
2x
dx
Solution: From y = Ae + (x + 2)e ,
2x
x
dy = Ae + e + 2(x + 2)e 2x
x
2x
dx
= Ae + (2x + 5)e 2x
x
Substitute this into the given equation,
dy

– y = Ae + (2x + 5)e 2x – [ Ae + (x + 2)e ]
x
x
2x
dx = (x + 3)e 2x

x
2x
Thus, y = Ae + (x + 2)e is a solution of the differential equation
dy – y = (x + 3)e .
2x
dx
From these two examples, note that the solution of the differential equation of the first order will contain one
arbitrary constant.
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04 STPM Math(T) T2.indd 128 28/01/2022 5:44 PM