The general solution of the differential equa | vedclass

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(D) Given,$\sin 2x \left( \frac{dy}{dx} - \sqrt{	an x} ight) - y = 0$ Rearranging the terms: $\frac{dy}{dx} - \frac{y}{\sin 2x} = \sqrt{	an x}$ Si

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$y\sqrt{\cot x} = x + c$

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(D) Given,$\sin 2x \left( \frac{dy}{dx} - \sqrt{	an x} ight) - y = 0$ Rearranging the terms: $\frac{dy}{dx} - \frac{y}{\sin 2x} = \sqrt{	an x}$ Since $\frac{1}{\sin 2x} = \csc 2x$,we have: $\frac{dy}{dx} - y \csc 2x = \sqrt{	an x}$ ....$(1)$ This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\csc 2x$ and $Q = \sqrt{	an x}$. Integrating Factor ($I$.$F$.) $= e^{\int P dx} = e^{\int -\csc 2x dx} = e^{-\frac{1}{2} \ln|	an x|} = e^{\ln(	an x)^{-1/2}} = \frac{1}{\sqrt{	an x}} = \sqrt{\cot x}$. The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$. Substituting the values: $y \sqrt{\cot x} = \int \sqrt{	an x} \cdot \sqrt{\cot x} dx + c$. Since $\sqrt{	an x} \cdot \sqrt{\cot x} = 1$,we get: $y \sqrt{\cot x} = \int 1 dx + c$. Therefore,$y \sqrt{\cot x} = x + c$.

The general solution of the differential equation,$\sin 2x \left( \frac{dy}{dx} - \sqrt{\tan x} \right) - y = 0$ is

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