2. SEPARABLE EQUATIONS
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Example (An initial value problem (IVP)). In an IVP, besides the di↵erential equation, we are given an initial condition that allows us to find a specific
solution rather than a general solution with a constant of integration.
dy
= 8x3e 2y , y(0) = 0
dx
Because of the initial condition, the only possible constant solution is y = 0,
but that is not a solution here. If it were, we would be done since there is only
a single solution to an IVP. We have
Z
Z
e2y dy = 8x3 dx =) e2y dy = 8x3 dx + K =)
1 2y
e = 2x4 + K =) e2y = 4x4 + 2K =)
2
ln e2y = ln(4x4 + 2K) =) 2y = ln(4x4 + C) =)
1
y = ln(4x4 + C)
2
This is the general solution. Then
1
y(0) = ln C = 0 =)
2
ln C = 0 =)
C=1
Thus the solution to the IVP is
p
1
4
y = ln(4x + 1) = ln 4x4 + 1
2