logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Differential Equations

The differential equation of the following equation $y=ae^{2x} +be^{-3 x} +ce^{x}$

$(a)\;\frac{7d^3 y}{dx^3}-\frac{7d^2y}{dx^2}+y=0 \\ (b)\;\frac{d^3 y}{dx^3}-7 \frac{d^2y}{dx^2}+6 \frac{d^2y}{dx^2}+y=0 \\ (c)\;\frac{d^3y}{dx^3}-\frac{7dy}{dx}+6y=0 \\ (d)\;\frac{d^2y}{dx^2}-6\frac{dy}{dx}+7=0 $

1 Answer

$y=ae^{2x}+3be^{-3x}+ce^{x}$
$\large\frac{dy}{dx}$$=2 a e^{2x}- 3 b e^{-3x} +ce^{x}$
$\large\frac{d^2y}{dx^2}$$=4 a e^{2x}+9 b e^{-3x} +ce^{x}$
$\large\frac{d^3y}{dx^3}$$=8 a e^{2x} -27 b e^{-3x} +ce^{x}$
$\large\frac{dy}{dx}$$-y=ae^{2x}- 4be^{-3x}$------(1)
$\large\frac{d^2y}{dx^2}$$-y=3 a e^{2x}+8 b e^{-3x} $--------(2)
$\large\frac{d^3y}{dx^3}$$-y=7 a e^{2x}- 28 b e^{-3x} $-------(3)
Equation (1) $\times$ 7
Equation (3) $\times$ 7
$\large\frac{7dy}{dx}-7y=\large\frac{d^3y}{dx^3} $$-y$
$\large\frac{d^3y}{dx^3}-\frac{7dy}{dx}$$+6y=0$
Hence c is the correct answer.
answered Feb 3, 2014 by meena.p
 

Related questions

Download clay6 mobile appDownload clay6 mobile app
...
X