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    Q15.    \frac{dy}{dx} - 3y \cot x = \sin 2x;\ y = 2\ when \ x = \frac{\pi}{2}

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Given equation is
\frac{dy}{dx} - 3y \cot x = \sin 2x;\ y = 2\ when \ x = \frac{\pi}{2}
This is  \frac{dy}{dx} + py = Q  type where p =-3\cot x and Q =\sin 2x
Now,
I.F. = e^{\int pdx}= e^{-3\cot xdx}= e^{-3\log|\sin x|}= \sin ^{-3}x= \frac{1}{\sin^3x}                     
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(\frac{1}{\sin^3 x}) =\int (\sin 2x\times\frac{1}{\sin^3 x})dx +C
\frac{y}{\sin^3 x} =\int (2\sin x\cos x\times\frac{1}{\sin^3 x})dx +C
\frac{y}{\sin^3 x} =\int (2\times \frac{\cos x}{\sin x}\times\frac{1}{\sin x})dx +C
\frac{y}{\sin^3 x} =\int (2\times\cot x\times cosec x)dx +C
\frac{y}{\sin^3 x} =-2cosec x +C
Now, by using boundary conditions we will find the value of C
It is given that  y = 2 when x= \frac{\pi}{2}
at   x= \frac{\pi}{2}
\frac{2}{\sin^3\frac{\pi}{2}} = -2cosec \frac{\pi}{2}+C\\ \\ 2 = -2 +C\\ C = 4
Now,
y= 4\sin^3 x-2\sin^2x\\
Therefore, the particular solution is  y= 4\sin^3 x-2\sin^2x\\

Posted by

Gautam harsolia

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