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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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If $y=e^{-x}(A\cos x+B\sin x)$, then $y$ is a solution of

\begin{array}{1 1}(A)\;\frac{d^2y}{dx^2}-2\frac{dy}{dx}=0 & (B)\;\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0\\(C)\;\frac{d^2y}{dx^2}+2\frac{dy}{dx}+2y=0 & (D)\;\frac{d^2y}{dx^2}+2y=0\end{array}

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  • The general solution of a differential equation is a relation between dependent and independent variables having $'n'$ arbitrary constants.
  • A general solution may have moer than one form, but the arbitrary constants must be the same
  • $\large\frac{d}{dx}$$(uv)=\large\frac{du}{dv}$$.v+\large\frac{dv}{dx}$$.u$
Given $y=e^{-x}(A \cos x +B \sin x)$
Let us expand the above equation as
$y=e^{-x} \;A \cos x +e^{-x} \;B \sin x$
Consider $ e^{-x}.A \cos x$
This in the product form , so let us apply product rule to differentiate this,
$\large\frac{d}{dx}$$(uv)=\large\frac{d}{dx}$$(u).v+\large \frac{d}{dx}$$(v).u$
Let $ u=e^{-x}$ then $ \large\frac{d}{dx}$$(u)=-e^{-x}$
Let $ v=A \cos x $ then $ \large\frac{d}{dx}$$(v)=-A \sin x$
Hence $ \large\frac{d}{dx}$$(uv)=-e^{-x}.A \cos x+(-A \sin x).e^{-x}$
$=-e^{-x}[A \cos x+A \sin x]$
Similarly, $e^{-x}.B \sin x=-e^{-x}[B \sin x-B \cos x]$
Therefore $ \large\frac{dy}{dx}$$=-e^{-x}[A \cos x+A \sin x+B \sin x-B \cos x]$
$ =-e^{-x}[A \cos x+B \sin x]-e^{-x}[A \sin x-B \cos x]$
But $e^{-x}[A \cos x+B \sin x]=y$
Hence $\large\frac{dy}{dx}$$=-y.-e^{-x}[A \sin x-B \cos x]$
Now differentiating w.r.t x we get,
$\large\frac{d^2y}{dx^2}=-\frac{dy}{dx}$$-[+e^{-x}[A \cos x+A \sin x+B \sin x+B \cos x]]$
$=-\large\frac{dy}{dx}$$-[e^{-x}(A \cos x+B \sin x)]+[e^{-x}(-A \sin x+B \cos x)]$
$\large\frac{d^2y}{dx^2}=\large\frac{-dy}{dx}-\bigg[\large\frac{dy}{dx}-y\bigg]-y $$\qquad => e^x[-A \sin x+B \cos x=\large\frac{dy}{dx}-y]$
Hence the correct option is $C$
answered May 23, 2013 by meena.p

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