If $\frac{\mathrm{d} y}{\mathrm{~d} x}=y+3, y+3\gt0$ and $y(0)=2$, then $y(\log 2)$ is equal to

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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 $\int \frac{\mathrm{d} y}{y+3}=\int \mathrm{d} x+\mathrm{c}$ $\begin{array}{ll} & \Rightarrow \log (y+3)=x+\mathrm{c} ...(i)\ & y=2 	ext { when } x=0 \ 	herefore \quad & \log (2+3)=0+\mathrm{c} \Rightarrow \mathrm{c}=\log 5 \ 	herefore \quad & \log (y+3)=x+\log 5...[From(i)] \ & \Rightarrow y+3=5 \mathrm{e}^x \ & \Rightarrow y=5 \mathrm{e}^x-3 \ 	herefore \quad & y(\log 2)=5 \mathrm{e}^{\log 2}-3=10-3=7 \end{array}$ ...[From (i)]

If $\frac{\mathrm{d} y}{\mathrm{~d} x}=y+3, y+3\gt0$ and $y(0)=2$, then $y(\log 2)$ is equal to

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