Math, asked by shahilaruby, 9 months ago

Ex.2. Solve (D^2- 8D+9) y = 8 sin 5x

Answers

Answered by MaheswariS
7

\underline{\textsf{Given:}}

\mathsf{(D^2-8D+9)y=8\,sin5x}

\underline{\textsf{To find:}}

\textsf{The general solution of the differential equation}

\underline{\textsf{Solution:}}

\textsf{Consider,}

\mathsf{D^2-8D+9=0}

\textsf{Characteristic equation is}

\mathsf{m^2-8m+9=0}

\mathsf{(m-9)(m+1)=0}

\mathsf{m=-1,9}

\textsf{Complementary function is}

\mathsf{Ae^{(-1)x}+Be^{9x}}

\textsf{Particular integral}

\mathsf{=\dfrac{8\;sin5x}{D^2-8D+9}}

\mathsf{=\dfrac{8\;sin5x}{-25-8D+9}}\;\;\;(D^2\implies\,-25)

\mathsf{=\dfrac{8\;sin5x}{-16-8D}}

\mathsf{=\dfrac{8\;sin5x}{-8(D+2)}}

\mathsf{=\dfrac{-sin5x}{D+2}}

\mathsf{=\dfrac{-sin5x}{D+2}{\times}\dfrac{D-2}{D-2}}

\mathsf{=\dfrac{-sin5x(D-2)}{D^2-4}}

\mathsf{=\dfrac{-sin5x(D-2)}{D^2-4}}

\mathsf{=\dfrac{D(-sin5x)+2\,sin5x}{D^2-4}}

\mathsf{=\dfrac{-5\,cos5x+2\,sin5x}{-25-4}}

\mathsf{=\dfrac{-5\,cos5x+2\,sin5x}{-29}}

\mathsf{=\dfrac{5\,cos5x-2\,sin5x}{29}}

\therefore\textsf{The general solution is}

\mathsf{y=}\,\textsf{Complementary function + Particular integral}

\implies\boxed{\mathsf{y=Ae^{-x}+Be^{9x}+\dfrac{5\,cos5x-2\,sin5x}{29}}}

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