The solution of the differential equation $\log \left(\frac{\mathrm{dy}}{\mathrm{d} x}\right)=9 x-6 \mathrm{y}+6$ is (given that $\mathrm{y}=1$ when $x=0$ )

Question

Aicharya · Accepted Answer

$\log \left(\frac{d y}{d x}ight)=9 x-6 y+6 \Rightarrow \frac{d y}{d x}=e^{9 x-6 y+6}$ $	herefore \frac{d y}{d x}=e^{9 x+6} \cdot e^{-6 y} \Rightarrow \frac{d y}{e^{-6 y}}=e^{9 x} \cdot e^{6} d x$ $	herefore \int e^{6 y} d y=e^{6} \int e^{9 x} d x$ $\frac{e^{6 y}}{6}=\frac{e^{6} e^{9 x}}{9}+C$ We have $x=0, y=1$ $	herefore \frac{e^{6}}{6}=\frac{e^{6}}{9}+C \Rightarrow C=\frac{e^{6}}{18}$ $	herefore \frac{e^{6 y}}{6}=\frac{e^{6} e^{9 x}}{9}+\frac{e^{6}}{18} \Rightarrow 3 e^{6 y}=2 e^{6+9 x}+e^{6}$

The solution of the differential equation $\log \left(\frac{\mathrm{dy}}{\mathrm{d} x}\right)=9 x-6…

Solution