Question

If $e^{\left(dy/dx\right)}=x+1$ given that when $x=0,y=3$, then minimum (local) value of $y=f\left(x\right)$ is?

Answer 1

$e^{dy/dx}=x+1$

$\Rightarrow \frac{dy}{dx}={\log }_e(x+1)=\ln (x+1)$

$\Rightarrow dy=\ln (x+1)dx$

$\Rightarrow y=\int \underset{u}{\ln (x+1)}\underset{v}{1dx}$ $\int uv=u\int v-\int u^1\int v$

$=\ln (x+1)x-\int \frac{1}{x+1}x$

$=\ln (x+1)x-\int \frac{x+1-1}{x+1}dx$

$=x\ln (x+1)-x+\ln (x+1)$

$y=(x+1)\ln (x+1)-x+C$

Given that when $x=0,y=3$

$\Rightarrow 3=1\left(0\right)-0+C$

$\Rightarrow C=3$

$\Rightarrow y=(x+1)\ln (x+1)-x+C$

Minimum occurs at $\frac{dy}{dx}=0$

$\Rightarrow \frac{(x+1)}{(x+1)}+\ln (x+1)-1=\ln (x+1)=0$

$\therefore x=0$

At $x=0;y=-0+C$

$y=3$

$\therefore$ Minimum value= $3$