What is the particular solution of the differential equation  2yy' = e^(x-y^2)  with y=-2 when x=4?

Oct 2, 2017

$y = - \sqrt{x}$

Explanation:

We have:

$2 y y ' = {e}^{x - {y}^{2}}$

This is a non-linear First Order ODE which we can write as:

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

Which in this form is separable, so we can "seperate the variables" , to get:

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

Which conveniently is directly integrable:

${e}^{{y}^{2}} = {e}^{x} + C$

Using the initial condition $y = - 2$ when $x = 4$:

${e}^{4} = {e}^{4} + C \implies C = 0$

Thus, the Particular Solution is:

${e}^{{y}^{2}} = {e}^{x}$
${y}^{2} = x$

Note that if we form an explicit solution for $y$ we get:

$y = \pm \sqrt{x}$

And with the initial conditions, only:

$y = - \sqrt{x}$

forms a valid solution.