Question

solve
a)xcosx cosy + Siny dy/do = 0
b) dy/dx = sin(x+y)
​

Answer 1

Step-by-step explanation:

We have,

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

$Let \: x + y = v \\ \implies1 + \frac{dy}{dx} = \frac{dv}{dx}$

Now,

$\frac{dv}{dx} - 1 = \sin(v)$

$\implies \frac{dv}{dx} =1 + \sin(v)$

$\implies \frac{dv}{1 + \sin(v) } = dx$

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

$\implies \int \frac{(1 - \sin(v) )dv}{1 - \sin^{2} (v) } = \int \: dx$

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

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

$\implies \int \sec^{2} (v)dv - \int \tan(v)\sec (v)dv = \int \: dx$

$\implies \tan (v) - \sec (v)= x + C$

$\tan(x + y) - \sec(x + y) = x +C$