Question

dy/dx = tan^2(x+y) solve it​

Answer 1

Step-by-step explanation:

We have,

$\frac{dy}{dx} = \tan^{2} (x + y)$

$Let\: x+y=v\\ \implies 1 + \frac{dy}{dx}=\frac{dv}{dx}$

$\frac{dv}{dx} - 1 = \tan^{2} (v)$

$\implies \frac{dv}{dx} = \sec ^{2} (v)$

$\implies \cos^{2} (v) dv = dx$

Integrating both sides,

$\implies \int\cos^{2} (v) dv = \int \: dx$

$\implies \: \int \frac{1 + \cos(2v) }{2} dv = \int \: dx$

$\implies \frac{1}{2} \int \: dv + \frac{1}{2} \int \: \cos(2v) dv = \int \: dx$

$\implies \frac{v}{2} - \sin(2v) = x + c$

$\implies \frac{x}{2} + \frac{y}{2} - 2\sin(x + y) \cos(x + y) = x + c$

$\implies \frac{y}{2} - 2 \sin(x + y) \cos(x + y) = \frac{x}{2} + c$

$\implies\:y-4\sin(x+y)\cos(x+y)=x+C$