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Q.
An integrating factor of the differential equation $ \frac{dy}{dx} = y \,tan\, x - y^{2} \,sec \,x$ is equal to :
Differential Equations
Solution:
Consider the differential equation
$ \frac{dy}{dx} = y \,tan\, x - y^{2} \,sec \,x$
Divide by $y^{2}$ on both the sides, we get
$\frac{1}{y^{2}}\left(\frac{dy}{dx}\right) = \frac{tan \,x}{y} - sec \,x ...\left(1\right)$
Let $\frac{1}{y} = z$
Diff both side, we get
$\frac{-1}{y^{2}} . \frac{dy}{dx} = \frac{dz}{dx}$
Put value of $\frac{1}{y^{2}} \frac{dy}{dx}$ in the equation $\left(1\right)$ , we get
$-\left(\frac{dz}{dx}\right) - \left(tan x\right) z = -sec \,x$
$\Rightarrow \quad\left(\frac{dz}{dx} \right) + \left(tan\, x\right) z = sec $
This is the linear diff equation in 'z' i.e.
This is of the form $\frac{dz}{dx} + P.z = Q$
then integrating factor $= e^{\int Pdx}$
$\therefore \quad$ In the given question
$I.F. = e^{\int\,tan \,x \,dx} = e^{log\left(sec\, x\right)} = sec \,x$