Question
If $y^{\prime}=y+1, y(0)=1$, then $y(\ln 2)=$

Answer

(c) : $y^{\prime}=y+1 \Rightarrow \frac{d y}{y+1}=d x$
$\Rightarrow \ln (y+1)=x+C$ (Integrating both sides)
Now, $y(0)=1 \Rightarrow C=\ln 2$
$\therefore \ln \left(\frac{y+1}{2}\right)=x \Rightarrow y+1=2 e^x \Rightarrow y=2 e^x-1$
So, $y(\ln 2)=2 e^{\ln 2}-1=4-1=3$

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