Question

Solve cos x dy/dx - y sin x = y^3 cos^2 x

Answer 1

Step-by-step explanation:

We have,

$\cos(x) \frac{dy}{dx} - y \sin(x) = {y}^{3} \cos^{2} (x)$

$\implies \frac{dy}{dx} - y \tan(x) = {y}^{3} \cos (x)$

$\implies {y}^{ - 3} \frac{dy}{dx} - y ^{ - 2} \tan(x) = \cos (x)$

$Let \: - {y}^{ - 2} = v \\ \implies{y}^{ - 3} \frac{dy}{dx} = \frac{1}{2} \frac{dv}{dx}$

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

$\implies \frac{dv}{dx} + 2v \tan(x) = 2 \cos (x)$

$I.F. = {e}^{ \int2 \tan(x) dx} = {e}^{2 ln | \sec(x) | } = {e}^{ ln \sec ^{2} (x) } = \sec^{2} ( x)$

$v. \sec^{2} (x) = \int2 \cos(x) \sec ^{2} (x) dx$

$\implies \: v. \sec^{2} (x) = \int2 \sec (x) dx$

$\implies \: v. \sec^{2} (x) = 2 ln | \sec(x) + \tan(x) | + c$

$\implies \: - {y}^{ - 2} . \sec^{2} (x) = 2 ln | \sec(x) + \tan(x) | + c$

Answer 2

Answer:

Given,

        $cos(x) \frac{dy}{dx} - ysin(x) = y^{3} cos^{2} x\\or, \frac{dy}{dx} - y tan(x) = y^{3} cos(x)\\ or, y^{-3} \frac{dy}{dx } - y^{-2} tan(x) = cos(x)$ ..................................(i)

Let,  

         $- y^{-2} = v$

 →    $y^{-3} \frac{dy}{dx} = \frac{1}{2} \frac{dv}{dx}$

Putting this value in equation (i), we will get,

 $\frac{dv}{dx} + 2vtan(x) = 2cos(x)$ ........................................................(ii)

Integrating factor(I.F.) = $e^\int 2tan(x) = e^2ln|secx| = sec^2 (x)$

$v .sec^2 x = \int2 cos(x) sec^2 x dx$

→ $v sec^2 x = \int 2 sec (x) dx$

→ $v. sec^2 x = 2 ln |sec(x) + tan(x)| + c$

→ $- \frac{1}{y^{2} } = 2ln|sec(x) + tan(x) | +c$