Question

What is the separable variable? Dy/dx = ye^-x^2, y(4) = 1?

Answer 1

Solution:

A differential equation is said to be separable variable if the variables can be separated.

Given,

$\frac{dy}{dx} = ye^{-x^2}$

$\Rightarrow \frac{dy}{y} = e^{-x^2}dx$

$\Rightarrow \int\frac{dy}{y}=\int e{^{-x^2}}dx$

But $\int e^{-x^2}dx$  is a non elementary integral. We can express it as a power series

$e^{-x^2}= 1 - \frac{x^2}{1!}+\frac{x^4}{2!}-\frac{x^6}{3!} +\frac{x^8}{4!}-..$

We can approximate it as

$e^{-x^2}= 1 - \frac{x^2}{1!}+\frac{x^4}{2!}-\frac{x^6}{3!}$

So differential equation transforms to

$\int\frac{dy}{y}=\int$ $(1 - \frac{x^2}{1!}+\frac{x^4}{2!}-\frac{x^6}{3!})dx$

 $\Rightarrow \ln |y|=x - \frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42} + C$

By initial condition y(4)=1

0 = $4 - \frac{4^3}{3}+\frac{4^5}{10}-\frac{4^7}{42} + C$

$\Rightarrow C=305.028$

So ln|y| = $x - \frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42} +305.028$

 $\Rightarrow y=e^{(x - \frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42} + 305.028)}$

This is the explicit solution of the given differential equation.