well, we have
y' = -1/2 e^(-2t) + 1/ln10 * 1/4 10^(4t) + c
y'(0)=0 means -1/2 + 1/(4 ln10) + c = 0, so
c = 1/2 - 1/(4 ln10)
Now follow the same logic to get y(t)
Solve the following differential equation with initial conditions:
y"=e^(-2t) + 10^(4t); y(0)=1, y'(0) = 0
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