Question

Solve $\frac{dy}{dx}=\frac{1}{cos(x+y)}$

Answer 1

$\frac{dy}{dx}=\frac{1}{cos(x+y)}$

Let $x+y=v$

$1+\frac{dy}{dx}=\frac{dv}{dx}$

$\frac{dy}{dx}=\frac{dv}{dx}-1$

The original differential equation becomes :

$\frac{dv}{dx}-1=\frac{1}{cosv}$

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

$\int \frac{\cos vdv}{1+\cos v}=\int dx$

$\int \frac{cosv+1-1}{1+cosv}dv=x+C$

$\int dv-\int \frac{1}{1+cosv}dv=x+C$

$v-\int \frac{1}{2co{s}^2\left(\frac{v}{2}\right)}dv=x+C$

$v-\frac{1}{2}\int se{c}^2\left(\frac{v}{2}\right)dv=x+C$

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

$x+y-\frac{1}{2}tan\frac{x+y}{2}=x+C$

$y-\frac{1}{2}tan\frac{x+y}{2}=C$