# How do you solve for the equation dy/dx=3x^2/e^2y that satisfies the initial condition f(0)=12?

How do you solve for the equation $\frac{dy}{dx}=\frac{3{x}^{2}}{{e}^{2}y}$ that satisfies the initial condition f(0)=12?
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Hofpoetb9
First of all, I think ther is a mistake in your writing, I think you wanted to write:
$\frac{dy}{dx}=\frac{3{x}^{2}}{{e}^{2y}}$
This is a separable differential equations, so:
${e}^{2y}dy=3{x}^{2}dx⇒\int {e}^{2y}dy=\int 3{x}^{2}dx⇒$
$\frac{1}{2}{e}^{2y}={x}^{3}+c$.
Now to find c let's use the condition: $f\left(0\right)=\frac{1}{2}$
$\frac{1}{2}{e}^{2\cdot \frac{1}{2}}={0}^{3}+c⇒c=\frac{1}{2}e$
So the solution is:
$\frac{1}{2}{e}^{2y}={x}^{3}+\frac{1}{2}e⇒{e}^{2y}=2{x}^{3}+e⇒2y=\mathrm{ln}\left(2{x}^{3}+e\right)⇒$
$y=\frac{1}{2}\mathrm{ln}\left(2{x}^{3}+e\right)$