Math, asked by riuu049, 3 months ago

CO3
Find the Particular Integral of y" + 2y' + 3y = Sin x is

Answers

Answered by MaheswariS
4

\textbf{Given:}

\mathsf{y''+2y'+3y=sinx}

\textbf{To find:}

\textsf{Particular integral of the differential equation}

\textbf{Solution:}

\textsf{Particular integral}

\mathsf{=\dfrac{sinx}{D^2+2D+3}}

\mathsf{=\dfrac{sinx}{-1+2D+3}}\;\;(D^2\implies\,-1)

\mathsf{=\dfrac{sinx}{2D+2}}

\mathsf{=\dfrac{sinx}{2(D+1)}{\times}\dfrac{D-1}{D-1}}

\mathsf{=\dfrac{(D-1)sinx}{2(D^2-1)}}

\mathsf{=\dfrac{D(sinx)-sinx}{2(D^2-1)}}\;\;(D^2\implies\;-1)

\mathsf{=\dfrac{cosx-sinx}{2(-1-1)}}

\mathsf{=\dfrac{cosx-sinx}{2(-2)}}

\mathsf{=\dfrac{cosx-sinx}{-4}}

\mathsf{=\dfrac{sinx-cosx}{4}}

\implies\boxed{\mathsf{Particular\;integral=\dfrac{sinx-cosx}{4}}}

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