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

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Answer 1 · 2017-10-02T13:43:50

$y = -sqrt(x)$

We have:

$2yy' = e^(x-y^2)$

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

$2y dy/dx = e^(x)e^(-y^2)$
$:. (2y)/e^(-y^2) dy/dx = e^(x)$
$:. 2ye^(y^2) dy/dx = e^(x)$

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

$int \ 2ye^(y^2) \ dy = int \ e^(x) \ dx$

Which conveniently is directly integrable:

$e^(y^2) = e^(x) + C$

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

$e^(4) = e^(4) + C => C = 0$

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:

$y = +- sqrt(x)$

And with the initial conditions, only:

$y = -sqrt(x)$

forms a valid solution.

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