$\displaystyle \displaystyle \frac{dy}{dx}=\cos(x+y)$

How can this be solved?

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- May 4th 2010, 05:06 PMJosh146Nonlinear 1st order ODE
$\displaystyle \displaystyle \frac{dy}{dx}=\cos(x+y)$

How can this be solved? - May 4th 2010, 06:18 PMpickslides
Would it help if $\displaystyle \cos(x+y)= \cos x \cos y - \sin x \sin y $ ??

- May 4th 2010, 06:25 PMKrizalid
put $\displaystyle t=x+y$ and the equation becomes separable.

- May 4th 2010, 06:42 PMRandom Variable
let $\displaystyle u = x+y $

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

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

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

integrate both sides

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

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

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