General Solution and Initial Value Problems Tutorial

Tutorial 2 - General Solution and Initial Value Problems




y
1. Use the transformation v = to find the general solution of the differential equation
x
dy y x2
= + 2
dx x 3y
[5 marks]

y
SOLUTION: Let v =
x
dy dv
such that y = vx and =x +v [1]
dx dx
Then the eqn becomes
dv 1 dv 1
x + v = v + 2, or x = 2 [1]
dx 3v dx 3v
Separating gives Z Z
2 dx
3v dv = [1]
x
Integrating gives
v 3 = ln(x) + C [1]
Hence
y = x(ln(x) + C)1/3 [1]


2. Solve the following initial value problem [5 marks]

dy
+ cos(x)y = 3x2 e− sin(x) ; y(0) = 1
dx

SOLUTION: This is linear and can be solved with an integrating factor
R
cos(x)d x
e = esin(x) [1]

Multiplying both sides by the integrating factor and integrating gives
Z
sin(x)
e y = 3x2 dx [1]

= x3 + C . [1]

Hence the general solution is

y = e− sin(x) x3 + C


[1]

Inserting the initial value gives
y(0) = C = 1

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