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Q.
The family of curves satisfying the differential equation $ \frac{dy}{dx}-2y\,tan \,x+y^{2}\,tan^{4} \,x = 0$ is
Differential Equations
Solution:
Divide the equation by $y^2$, we get
$y^{-2} \frac{dy}{dx}-\left(2\, tan \,x\right)y^{-1} = -tan^{4} \,x$ [see the Bernoulli’s equation]
Put $y^{-1} = z \Rightarrow -y^{-2} \frac{dy}{dx} = \frac{dz}{dx}$
Hence, $-\frac{dz}{dx}-\left(2\, tan\, x\right).z = -tan^{4}\, x $
$\Rightarrow \frac{dz}{dx} + \left(2 \,tan\, x\right)z = tan^{4} \,x$
Which is linear in z, Integrating factor,
$I.F. = e^{\int \,2\, tan \,xdx} = e^{2\, log \left|sec x\right| } = sec^{2} \,x$
The solution is $z\left(sec^{2}\, x\right) = \int \,\left(tan^{4} \,x\right) sec^{2} \,xdx + a$
$y^{-1}\left(sec^{2}\, x\right) = \frac{1}{2} tan^{5} \,x +a \Rightarrow 5sec^{2} x = y \left(tan^{5}\, x +c\right),\, C = 5a$