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  • Find a particular solution of the differential equation dy/dx + y cot x = 4x cosec x (x not equal to 0), given that y = 0 when x = pi/2

Q13.    Find a particular solution of the differential equation \frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0), given that y = 0 \ \textup{when}\ x = \frac{\pi}{2}.

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Given equation is
\frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0)
This is  \frac{dy}{dx} + py = Q  type where p =\cot x and Q = 4xcosec xQ = 4x \ cosec x
Now,
I.F. = e^{\int pdx}= e^{\int \cot xdx}= e^{\log |\sin x|}= \sin x                     
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(\sin x ) =\int (\sin x\times 4x \ cosec x)dx +C
y(\sin x) =\int(\sin x\times \frac{4x}{\sin x}) +C\\ \\ y(\sin x) = \int 4x + C\\ y\sin x= 2x^2+C
Now, by using boundary conditions we will find the value of C
It is given that  y = 0 when x= \frac{\pi}{2}
at   x= \frac{\pi}{2}
0.\sin \frac{\pi}{2 } = 2.\left ( \frac{\pi}{2} \right )^2+C\\ \\ C = - \frac{\pi^2}{2}
Now, put the value of C
y\sin x= 2x^2-\frac{\pi^2}{2}
Therefore, the particular solution is  y\sin x= 2x^2-\frac{\pi^2}{2}, (sinx\neq0) 

Posted by

Gautam harsolia

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