If $\frac{\mathrm{d} y}{\mathrm{~d} x}=y+3$ and $y(0)=2$, then $y(\log 2)=$

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=y+3 \ & \Rightarrow \frac{\mathrm{d} y}{y+3}=\mathrm{d} x \end{aligned}$ Integrating on both sides, we get $\begin{array}{ll} & \int \frac{\mathrm{d} y}{y+3}=\int \mathrm{d} x+\mathrm{c} \ \Rightarrow & \log (y+3)=x+\mathrm{c} \ & y=2 	ext { when } x=0 \ 	herefore \quad & \log (2+3)=0+\mathrm{c} \Rightarrow \mathrm{c}=\log 5 \ 	herefore \quad & \log (y+3)=x+\log 5 \ & \Rightarrow y+3=5 \mathrm{e}^x \ \Rightarrow y=5 \mathrm{e}^x-3 & \ 	herefore \quad & y(\log 2)=5 \mathrm{e}^{\log 2}-3=10-3 \ & =7 \end{array}$

If $\frac{\mathrm{d} y}{\mathrm{~d} x}=y+3$ and $y(0)=2$, then $y(\log 2)=$

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