Initial value problem through origin

Question

$\frac{dz}{dt}=8t*e^z$, Through the origin

I have never done an initial value problem before, but I took it to mean that it gave me the initial value of the differential equation (0, 0) and that I should find the general value of the differential equation from it. Please correct me if I am wrong about this.

I started off by separating the variables:

$\frac{dz}{e^z}=8t*dt$

I then integrated:

$-e^{-z}=4t^2+c$

After dividing by -1 and taking the natural log I have:

$z=-ln(-4t^2+c)$

Here is where my problem is, no matter what I plug in for c I cannot get the function to equal 0 when t = 0. Even for another value of t it does not seem to possible to get 0 with this differential equation.

Answer 1

You solved the ODE correctly. We have $$z(t) = -\ln(-4t^2+c) \implies z(0) = -\ln(c) \implies \ln(c) = 0 \implies c = 1,$$ because $\ln 1 = 0.$ Ok?