Mathematics - Differential Equation Question with Solution | TestHub

We have,

$\frac{dy}{dx}=\sin \left(x+y\right)+\cos \left(x+y\right)$

Put $x+y=v\Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}$

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

$\Rightarrow \frac{dv}{dx}=\sin v+\cos v+1$

$\Rightarrow \frac{dv}{\sin v+\cos v+1}=dx$

$\Rightarrow \int \left(\frac{1}{\sin v+\cos v+1}\right)dv=\int dx$

$\Rightarrow \int \frac{1}{\left\{{\displaystyle \frac{2\tan \left({\displaystyle \frac{v}{2}}\right)}{1+{\tan }^2\left({\displaystyle \frac{v}{2}}\right)}}\right\}+\left\{{\displaystyle \frac{1-{\tan }^2\left({\displaystyle \frac{v}{2}}\right)}{1+{\tan }^2\left({\displaystyle \frac{v}{2}}\right)}}\right\}+1}dv=\int dx$

$\Rightarrow \int \left[\frac{\left\{1+{\tan }^2\left({\displaystyle \frac{v}{2}}\right)\right\}}{{\displaystyle 2}{\displaystyle \tan }{\displaystyle \left(\frac{{\displaystyle v}}{{\displaystyle 2}}\right)}+{\displaystyle 1}{\displaystyle -}{\displaystyle {{\displaystyle \tan }}^2}{\displaystyle \left(\frac{{\displaystyle v}}{{\displaystyle 2}}\right)}+1+{\tan }^2\left({\displaystyle \frac{v}{2}}\right)}\right]dv=x+C$

$\Rightarrow \frac{1}{2}\int \left[\frac{{\sec }^2\left({\displaystyle \frac{v}{2}}\right)}{{\displaystyle \tan \left(\frac{{\displaystyle v}}{{\displaystyle 2}}\right)+1}}\right]dv=x+C$

$\Rightarrow \log \left(\tan \left(\frac{{\displaystyle v}}{{\displaystyle 2}}\right)+1\right)=x+C$

$\Rightarrow \log \left(\tan \left(\frac{{\displaystyle x+y}}{{\displaystyle 2}}\right)+1\right)=x+C$