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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Which of the following is the general solution of $\large\frac{d^2y}{dx^2}$$-2\large\frac{dy}{dx}$$+y=0$

\[(A)\;y=(Ax+B)e^x \quad (B)\;y=(Ax+B)e^{-x}\quad(C)\;y=Ae^x+Be^{-x} \quad (D)\;y=A\cos x+B\sin x\]

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1 Answer

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Toolbox:
  • If the given equation has $'n'$ arbitary constants then its differential equation will be of $n^{th}$ order.
  • $\large\frac{d}{dx}$$(uv)=\large\frac{du}{dx}.$$v+u.\large\frac{dv}{dx}$
Given $\large\frac{d^2y}{dx^2}$$-2\large\frac{dy}{dx}$$+y=0$
Consider the equation $y=(Ax+B)e^x $
On expanding we get,
$y=Axe^x+Be^x$
Differentiate w.r.t x on both sides,
$\large\frac{dy}{dx}$$=A\bigg[\large\frac{d}{dx}$$(x).e^x+\large\frac{d}{dx}$$(e^x).x\bigg]$$+Be^x$
=>$\large\frac{dy}{dx}$$=A[e^x+xe^x]+Be^x$
=>$\large\frac{dy}{dx}$$=Ae^x+Axe^x+Be^x \qquad (But:(Ax+B)e^x=y)$
$=Ae^x+y$
$\large\frac{dy}{dx}$$=Ae^x+y=>Ae^x=\large\frac{dy}{dx}$$-y$
Again differentiating we get,
$\large\frac{d^2y}{dx^2}$$=Ae^x+\large\frac{dy}{dx}$
Substituting for $Ae^x$
$\large\frac{d^2y}{dx^2}=\bigg(\large\frac{dy}{dx}$$-y\bigg)+\large\frac{dy}{dx}$
$=2 \large\frac{dy}{dx}$$-y$
Therefore $\large\frac{d^2y}{dx^2}$$-2\large \frac{dy}{dx}$$+y=0$
Hence the correct option is $A$

 

answered May 20, 2013 by meena.p
 

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