# Thread: Differential Equation Substitution

1. ## Differential Equation Substitution

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

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

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

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

and then find the general solution for y(x)

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

Am I on the right track, and where do I go from here?

2. Originally Posted by jwade456
Hey guys I'm a bit stuck on this problem, any help would be greatly appreciated!

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

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

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

When I make that substitution, I get:

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

The DE however, looks like a Riccati equation:

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

So for:

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

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

In that case I get:

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

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