Question

Solution of $e^{\frac{dy}{dx}}=x$ when $x=1$ and $y=0$ is

• A. $y=x(logx-1)+4$
• B. $y=x(logx-1)+3$
• C. $y=x(logx+1)+1$
• D. $y=x(logx-1)+1$

Answer 1

Answer: (D)

Given, $e^{dy/dx}=x$
Taking logarithm on both sides, we get
$log{e}^{dydx}=logx$
$\frac{dy}{dx}=$ = log {\,}x
$dy=logxdx$
On integrating , we get
$\int dy=\int logxdx$
$=logx\int 1dx=\int \left[\frac{d}{dx}logx\int 1dx\right]dx$
$=xlogx-\int \frac{1}{x}\times xdx$
$=xlogx-\int dx$
$=xlogx-x$
$y=x(logx-l)+C\dots (i)$
when $x=1$ and $y=0$
$\Rightarrow 0=1(log1-1)+C$
$\Rightarrow 0=(0-1)+C$
$\Rightarrow C=1$
$\therefore$ Eq. (i) becomes
$y=x(logx-1)+1$