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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation

$y=x\;\sin 3x$ $\:$:$\:$ $\frac{d^2y}{dx^2}+9y-6\;\cos 3x=0$
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  • $\large\frac{d(e^x)}{dx }$$= e^x$; $\large\frac{d(x)}{dx}$$ = 1$; $\large\frac{d(constant\; term )} {dx}= $$0$. The product rule of differentiation states $uv = uv' + vu'$
Step 1:
Given $y = x\sin 3x$
Differentiating on both sides we get,
$\large\frac{dy}{dx} $$= x. (3\cos3x) + (\sin3x).1$
$\large\frac{dy}{dx} $$= 3x\cos 3x + \sin 3x$
differentiating again we get
$\large\frac{d^2y}{dx^2 }$$= 3x.(- 3\sin 3x) + 3x(\cos 3x) + (3\cos 3x)$
$\large\frac{d^2y}{dx^2} $$= -9x\sin 3x + 3x\cos 3x +3\cos 3x$
Step 2:
Substituting in the given solution for $\large\frac{d^2y}{dx^2},\large\frac{dy}{dx}$ and $y$ we get
$-9x\sin 3x + 3x\cos 3x + 3\cos 3x + 9(x\sin 3x) - 6\cos 3x =0$
Hence the given function is the solution for the given differential equation.
answered Jul 30, 2013 by sreemathi.v

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