The particular solution of $\log \left(\frac{d y}{d x}\right)=3 x+4 y$ at $x=y=0$ is

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Aicharya · Accepted Answer

$\begin{aligned} & \log \left(\frac{d y}{d x}ight)=3 x+4 y 	ext { and } x=y=0 \ & \Rightarrow \frac{d y}{d x}=e^{3 x+4 y} \ & \Rightarrow \int e^{-4 y} d y=\int e^{3 x} d x \ & \Rightarrow \frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}+C\end{aligned}$ Putting $x=0$ and $y=0$ we get $C=\frac{-7}{12}$ $\begin{aligned} & \frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}-\frac{7}{12} \ & \Rightarrow 3 e^{-4 y}=-4 e^{3 x}+7 \ & \Rightarrow 3 e^{-4 y}+4 e^{3 x}=7\end{aligned}$

The particular solution of $\log \left(\frac{d y}{d x}\right)=3 x+4 y$ at $x=y=0$ is

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