Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

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}

Can you answer this question?

1 Answer

0 votes
  • 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App