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)

$y = x(logx - 1)+ 1$

Answer 2

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