If $\frac{d y}{d x}=y+3$ and $y(0)=2$, then $y(\ln 2)$ is equal to

7

We have,

$\frac{d y}{d x}=y+3$

$\Rightarrow \frac{1}{y+3} d y=d x$

$\Rightarrow \int \frac{1}{y+3} d y=\int 1 . d x$

$\Rightarrow \log (y+3)=x+C$            ......(i)

It is given that $y(0)=2$ i.e. $y=2$ when $x=0$

∴   $\log 5=C$            [Putting y = 2, x = 0 in (i)]

Substituting the value of $C$ in (i), we get

$\log (y+3)=x+\log 5$

$\Rightarrow y+3=5 e^x$

Putting $x=\ln 2$, we get

$y+3=5 e^{\log 2} \Rightarrow y+3=10 \Rightarrow y=7$