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True-or-False: Differential equation representing the family of curves $y=e^x(A\cos x+B\sin x)$ is $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$.

$\begin{array}{1 1} True \\ False \\ Cannot\;be\;determined \end{array}$

1 Answer

Toolbox:
  • If the given equation has $n$ arbitrary constants then its differential eqaution will be of $n^{th}$ order.
  • Product rule: $\large\frac{d}{dx}$$(uv)=u. \large\frac{dv}{dx}$$+v.\large\frac{du}{dx}$
  • $\int e^{-ax}dx=-\large\frac{1}{a}$$ e^{ax}$
$y=e^x(A \cos x+B \sin x)$
This can be written as $y e^{-x}=A\cos x+B \sin x$
$ye^{-x}$ is a product of two functions, Hence it can be differentiated by using product rule
$\large\frac{d}{dx}$$(uv)=u. \large\frac{dv}{dx}$$+v.\large\frac{du}{dx}$
Let $u=y,$ then $\large\frac{du}{dx}=\frac{dy}{dx}$ and let $v=e^{-x}$ then $\large\frac{dv}{dx}=-e^{-x}$
$\large\frac{d}{dx}$$(uv)=-y.e^{-x}+e^{-x}.\large\frac{dy}{dx}$
Hence differentiating $ye^{-x}=A \cos x+B \sin x$
$-ye^{-x}+e^{-x}.\large\frac{dy}{dx}$$=-A \cos x+B \sin x$
Now again differentiating similarly as above by applying the product rule. we get,
$-\bigg[y.(-e^{-x})+e^{-x}.\large\frac{dy}{dx}\bigg]$$+e^{-x}.\large\frac{d^2y}{dx^2}+\frac{dy}{dx}$$.(-e^{-x})=-A \cos x-B \sin x$
$=>ye^{-x}-e^{-x}.\large\frac{dy}{dx}$$+e^{-x}.\large\frac{d^2y}{dx^2}-\frac{dy}{dx}$$.e^{-x}=-(A \cos x+B \sin x)$
$=>e^{-x}\bigg[\large\frac{d^2y}{dx^2}$$-2\large\frac{dy}{dx}$$+y\bigg]=-(A \cos x+B \sin x)$
$=>\large\frac{d^2y}{dx^2}$$-2\large\frac{dy}{dx}$$+y=-e^x(A \cos x+B \sin x)$
But $e^x(A \cos x+B \sin x)=y$
$\large\frac{d^2y}{dx^2}$$-2\large\frac{dy}{dx}$$+2y=0$
Hence the statement
Differential equation representing the family of curves $y=e^x(A\cos x+B\sin x)$ is $\large\frac{d^2y}{dx^2}$$-2\large\frac{dy}{dx}$$+2y=0$. is $ True$
answered May 23, 2013 by meena.p
 

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