# Nonlinear 1st order ODE

• May 4th 2010, 05:06 PM
Josh146
Nonlinear 1st order ODE
$\displaystyle \frac{dy}{dx}=\cos(x+y)$

How can this be solved?
• May 4th 2010, 06:18 PM
pickslides
Would it help if $\cos(x+y)= \cos x \cos y - \sin x \sin y$ ??
• May 4th 2010, 06:25 PM
Krizalid
put $t=x+y$ and the equation becomes separable.
• May 4th 2010, 06:42 PM
Random Variable
let $u = x+y$

then $y= u-x$ and $\frac{dy}{dx} = \frac{du}{dx} -1$

so $\frac{du}{dx} -1 = \cos u$

$\frac{du}{1+\cos u} = dx$

integrate both sides

$\tan\Big(\frac{u}{2}\Big) = x + C$ (the substitution $v = \tan\Big(\frac{u}{2}\Big)$ transforms the integral on the left into $\int dv$ )

then $u = 2 \arctan (x+C)$

and finally $y = 2 \arctan(x+C)-x$
• May 4th 2010, 06:53 PM
Krizalid
$\frac{1}{1+\cos u}=\frac{1-\cos u}{\sin ^{2}u}=\csc ^{2}(u)-\cot (u)\csc (u),$ faster to integrate. :D