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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?

$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