Question

Solve the differential equation $e^{dy/dx}=x+1$ given thet when $x=0,y=3.$

• A. $y=(x+1)\log (x+1)-x+3$
• B. $y=(x+1)\log (x+1)-x-3$
• C. $y=(x+1)\log (x+1)+x+3$
• D. None of these

Answer 1

The correct option is A $y=(x+1)\log (x+1)-x+3$

Given, $e^{\frac{dy}{dx}}=x+1$

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

Integrating both sides
$y=\int \log (x+1)dx.$

Put $x+1=t\Rightarrow dx=dt$

$y=\int \log tdt=t\log t-t+c$
$\Rightarrow y=(x+1)\log (x+1)-(x+1)+c$
Put $x=0,y=3\Rightarrow c=4$
$\therefore y=(x+1)\log (x+1)-x+3$