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The differential equation for $y=A\cos\alpha x+B\sin\alpha x,$where A and B are arbitrary constants is\begin{array}{1 1}(A)\;\frac{d^2y}{dx^2}+x^2y=0 & (B)\;\frac{d^2y}{dx^2}+\alpha^2y=0\\(C)\;\frac{d^2y}{dx^2}+y=0 & (D)\;\frac{d^2y}{dx^2}-y=0\end{array}

1 Answer

  • The general solution of a differential equation is a relation between the dependent and the independent variables having $'n'$ arbitary constants.
  • A general solution may have more than one form, but the arbitary constants must be the same
$y=A \cos \alpha x+B \sin \alpha x$
Let us differentiate this w.r.t x on both sides
$\Rightarrow \large \frac{dy}{dx}$$=-\alpha A \sin \alpha x+\alpha B \cos \alpha x$
Again differentiating w.r.t x we get
$\Rightarrow \large \frac{d^2y}{dx^2}$$=-\alpha ^2 A \cos \alpha x-\alpha ^2B \sin \alpha x$
$\large \frac{d^2y}{dx^2}$$=-\alpha ^2(A \cos \alpha x+B\sin \alpha x)$
But $ A \cos \alpha x+ B \sin \alpha x=y$
$\therefore \large \frac{d^2y}{dx^2}$$=-\alpha ^2y$
$\Rightarrow \large \frac{d^2y}{dx^2}$$+\alpha ^2y=0$
Hence the correct option is B
answered May 14, 2013 by meena.p
edited Oct 1, 2014 by rvidyagovindarajan_1

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