Question

The family of curves satisfying the differential equation $\frac{dy}{dx}-2ytanx+y^{2}ta{n}^{4}x=0$ is

• A. $yse{c}^{2}x=5ta{n}^{2}x+c$
• B. $ysi{n}^{5}x+5co{s}^{2}x$
• C. $5se{c}^{2}x=y\left(ta{n}^{5}x+c\right)$
• D. None of these

Answer 1

Answer: (C)

Divide the equation by $y^2$, we get
$y^{-2}\frac{dy}{dx}-\left(2tanx\right)y^{-1}=-ta{n}^{4}x$ [see the Bernoulli’s equation]
Put $y^{-1}=z\Rightarrow -y^{-2}\frac{dy}{dx}=\frac{dz}{dx}$
Hence, $-\frac{dz}{dx}-\left(2tanx\right).z=-ta{n}^{4}x$
$\Rightarrow \frac{dz}{dx}+\left(2tanx\right)z=ta{n}^{4}x$
Which is linear in z, Integrating factor,
$I.F.=e^{\int 2tanxdx}=e^{2log|secx|}=se{c}^{2}x$
The solution is $z\left(se{c}^{2}x\right)=\int \left(ta{n}^{4}x\right)se{c}^{2}xdx+a$
$y^{-1}\left(se{c}^{2}x\right)=\frac{1}{2}ta{n}^{5}x+a\Rightarrow 5se{c}^{2}x=y\left(ta{n}^{5}x+c\right),C=5a$