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

General solution of differential equation : $x^2e^y dy - (x-1) e^x dx=0$

$(a)\;e^x=\frac{e^y}{x}+c \\ (b)\;e^y=\frac{e^x}{x}+c \\ (c)\;e^x= \frac{e^y}{y}+c \\ (d)\;e^y= \frac{e^x}{y}+c $
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1 Answer

$x^2e^y dy -(x-1) ^x edx=0$
$e^y dy =\large\frac{x-1}{x^2}$$e^x dx$
$d(e^y) =e^x \bigg[ \large\frac{1}{x}- \frac{1}{x^2}\bigg]$$dx$
$d(e^y)= d \bigg( \frac{e^x}{x}\bigg)$
$e^y =\large\frac{e^x}{x}$$+c$
Hence b is the correct answer.
answered Feb 3, 2014 by meena.p
 

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