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  • Find general solution: dy/dx + 2y + sin x

Find the general solution:

    Q1.    \frac{dy}{dx} + 2y = \sin x

Answers (1)

best_answer

Given equation is
\frac{dy}{dx} + 2y = \sin x
This is  \frac{dy}{dx} + py = Q  type where p = 2 and Q = sin x  
Now,
I.F. = e^{\int pdx}= e^{\int 2dx}= e^{2x}
Now, the solution of given differential equation is given by relation
Y(I.F.) =\int (Q\times I.F.)dx +C
Y(e^{\int 2x }) =\int (\sin x\times e^{\int 2x })dx +C
Let  I =\int (\sin x\times e^{\int 2x })
I = \sin x \int e^{2x}dx- \int \left ( \frac{d(\sin x)}{dx}.\int e^{2x}dx \right )dx\\ \\ I = \sin x.\frac{e^{2x}}{2}- \int \left ( \cos x.\frac{e^{2x}}{2} \right )\\ \\ I = \sin x. \frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x\int e^{2x}dx- \left ( \frac{d(\cos x)}{dx}.\int e^{2x}dx \right ) \right )dx\\ \\ I = \sin x\frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x.\frac{e^{2x}}{2}+ \int \left ( \sin x.\frac{e^{2x}}{2} \right ) \right )\\ \\ I = \sin x\frac{e^{2x}}{2}-\frac{1}{2}\left ( \cos x.\frac{e^{2x}}{2}+\frac{I}{2} \right ) \ \ \ \ \ \ \ \ \ \ \ (\because I = \int \sin xe^{2x})\\ \\ \frac{5I}{4}= \frac{e^{2x}}{4}\left ( 2\sin x-\cos x \right )\\ \\ I = \frac{e^{2x}}{5}\left ( 2\sin x-\cos x \right )
Put the value of I in our equation
Now, our equation become 
Y.e^{x^2 }= \frac{e^{2x}}{5}\left (2 \sin x-\cos x \right )+C
Y= \frac{1}{5}\left (2 \sin x-\cos x \right )+C.e^{-2x}
Therefore, the general solution is Y= \frac{1}{5}\left (2 \sin x-\cos x \right )+C.e^{-2x}
 

Posted by

Gautam harsolia

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