Differential Equation Substitution

• October 2nd 2008, 03:35 AM
Differential Equation Substitution
Hey guys I'm a bit stuck on this problem, any help would be greatly appreciated!

For, $\frac{dy}{dx}=\frac{2cos^{2}x-sin^{2}x+y^2}{2cos^{2}x}$

Substitute, $y(x)=sinx+\frac{1}{u(x)}$

to get, $\frac{du}{dx}=-utanx-\frac{1}{2}secx$

and then find the general solution for y(x)

So far I've got to $u^{2}=\frac{2cosx}{sin^{2}x+y^{2}}$

Am I on the right track, and where do I go from here?
• October 2nd 2008, 07:06 AM
shawsend
Quote:

Hey guys I'm a bit stuck on this problem, any help would be greatly appreciated!

For, $\frac{dy}{dx}=\frac{2cos^{2}x-sin^{2}x+y^2}{2cos^{2}x}$

Substitute, $y(x)=sinx+\frac{1}{u(x)}$

to get, $\frac{du}{dx}=-utanx-\frac{1}{2}secx$

When I make that substitution, I get:

$u'=-1/2\sec^2(x)-\sec(x)\tan(x)u+(\cos(x)-1)u^2$

The DE however, looks like a Riccati equation:

$y'+Q(x)y+R(x)y^2=P(x)$

So for:

$\frac{dy}{dx}=1-1/2\tan^2(x)+1/2\sec^2(x)y^2$

The standard approach for these is to let $y=\frac{u'}{Ru}$

In that case I get:

$u''+2\tan(x)u'-1/8\sec^2(x)(1/2\tan^2(x)-1)u=0$

Second order I know, but we've changed a non-linear equation to a linear one.