# Use substitution to solve a differential equation

• Dec 19th 2011, 01:18 AM
Punch
Use substitution to solve a differential equation
By using the substitution $v=x-y$, solve the differential equation $\frac{dy}{dx}+[1+(x-y)^2]cos^2x=sin^2x$, expressing $y$ in terms of $x$.

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

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

How do I continue to remove the x in the trigo?
• Dec 19th 2011, 01:42 AM
alexmahone
Re: Use substitution to solve a differential equation
Quote:

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

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

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

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

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

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

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

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

Can you proceed?
• Dec 19th 2011, 02:31 AM
Punch
Re: Use substitution to solve a differential equation
Quote:

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

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

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

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

Can you proceed?

Yeap, thanks!