Question

The particular solution of the differential equation $\frac{dy}{dx}=\ln (x+1),x>-1,y\left(0\right)=3,$ is given by

• A. $y=(1-x)\ln (1-x)+x+3$
• B. $y=\frac{\ln (1+x)}{1+x}-x+3$
• C. $y=(x+1)\ln (x+1)-x+3$
• D. $y=\frac{\ln (1-x)}{{(1+x)}^2}+x+3$

Answer 1

The correct option is C $y=(x+1)\ln (x+1)-x+3$

Given,
$\frac{dy}{dx}=\ln (x+1)$
$\Rightarrow \int dy=\int \ln (x+1)dx$
$\Rightarrow y=(x+1)\ln (x+1)-x+c$
When $x=0,y=3$ gives $c=3$
Hence the solution is $y=(x+1)\ln (x+1)-x+3$