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Find the curve $y= f(x) $ through the origin for which $y''=y'$ and tangent at origin is $y=x$

$(a)\;y=e^x \\ (b)\;y=x-1 \\ (c)\;y=e^x-1 \\ (d)\;y=e^{x-1} $
Can you answer this question?
 
 

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$y'= \large\frac{dy}{dx}$$=t$
$y''= \large\frac{dt}{dx}$
$\large\frac{dt}{dx}$$=t$
$\log t= x+ \log c$
$t= ce^x$
$\large\frac{dy}{dx} $$=ce^x$
$dy= ce^x dx$
$y= ce^x+c'$
(0,0) satisfied
So, $=c+c'$
Slope of $y=x$ is 1
So, $\large\frac{dy}{dx}$$=ce^x=1$
at $(x=0,y=0)$
$c=1$
$c'=-1$
$y=e^x-1$
Hence c is the correct answer.
answered Feb 6, 2014 by meena.p
 

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