Question

50. Solve dy/dx
= y^2tan 2x, given that y = 2 when x = 0.​

Answer 1

Step-by-step explanation:

We have,

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

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

Integrating both sides,

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

$\implies - \frac{1}{y} = \frac{ \sec^{2} (2x) }{2} + c$

Since, y=2 when x=0, then,

$\implies - \frac{1}{2} = \frac{1}{2} + c$

$\implies \: c = - \frac{1}{4}$

so, required equation

$\implies - \frac{1}{y} = \frac{ \sec^{2} (2x) }{2} - \frac{1}{4}$

$\implies - \frac{4}{y} = 2 \sec^{2} (2x) - 1$

$\implies \: y = \frac{4}{1 - 2 \sec^{2} (2x) }$