# Use substitution to solve a differential equation

• Dec 19th 2011, 12:18 AM
Punch
Use substitution to solve a differential equation
By using the substitution $\displaystyle v=x-y$, solve the differential equation $\displaystyle \frac{dy}{dx}+[1+(x-y)^2]cos^2x=sin^2x$, expressing $\displaystyle y$ in terms of $\displaystyle x$.

$\displaystyle \frac{dv}{dx}=1-\frac{dy}{dx}$

$\displaystyle 1-\frac{dv}{dx}+[1+v^2]cos^2x=sin^2x$

How do I continue to remove the x in the trigo?
• Dec 19th 2011, 12:42 AM
alexmahone
Re: Use substitution to solve a differential equation
Quote:

Originally Posted by Punch
By using the substitution $\displaystyle v=x-y$, solve the differential equation $\displaystyle \frac{dy}{dx}+[1+(x-y)^2]cos^2x=sin^2x$, expressing $\displaystyle y$ in terms of $\displaystyle x$.

$\displaystyle \frac{dv}{dx}=1-\frac{dy}{dx}$

$\displaystyle 1-\frac{dv}{dx}+[1+v^2]cos^2x=sin^2x$

How do I continue to remove the x in the trigo?

$\displaystyle \frac{dv}{dx}=1+\cos^2x+v^2\cos^2x-\sin^2x$

$\displaystyle =2\cos^2x+v^2\cos^2x$

$\displaystyle =\cos^2x(v^2+2)$

$\displaystyle \frac{dv}{v^2+2}=\cos^2xdx$

Can you proceed?
• Dec 19th 2011, 01:31 AM
Punch
Re: Use substitution to solve a differential equation
Quote:

Originally Posted by alexmahone
$\displaystyle \frac{dv}{dx}=1+\cos^2x+v^2\cos^2x-\sin^2x$

$\displaystyle =2\cos^2x+v^2\cos^2x$

$\displaystyle =\cos^2x(v^2+2)$

$\displaystyle \frac{dv}{v^2+2}=\cos^2xdx$

Can you proceed?

Yeap, thanks!