# How do you solve for the equation dy/dx=(3x^2)/(e^2y) that satisfies the initial condition f(0)=1/2?

Feb 22, 2015

The answer is: $y = \frac{1}{2} \ln \left(2 {x}^{3} + e\right)$.

First of all, I think ther is a mistake in your writing, I think you wanted to write:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2}}{e} ^ \left(2 y\right)$.

This is a separable differential equations, so:

${e}^{2 y} \mathrm{dy} = 3 {x}^{2} \mathrm{dx} \Rightarrow \int {e}^{2 y} \mathrm{dy} = \int 3 {x}^{2} \mathrm{dx} \Rightarrow$

$\frac{1}{2} {e}^{2 y} = {x}^{3} + c$.

Now to find $c$ let's use the condition: $f \left(0\right) = \frac{1}{2}$

$\frac{1}{2} {e}^{2 \cdot \frac{1}{2}} = {0}^{3} + c \Rightarrow c = \frac{1}{2} e$.

So the solution is:

$\frac{1}{2} {e}^{2 y} = {x}^{3} + \frac{1}{2} e \Rightarrow {e}^{2 y} = 2 {x}^{3} + e \Rightarrow 2 y = \ln \left(2 {x}^{3} + e\right) \Rightarrow$

$y = \frac{1}{2} \ln \left(2 {x}^{3} + e\right)$.