Question

Find the particular solution of the differential equation $\log \left(\frac{dy}{dx}\right)=3x+4y$, given that $y=0$ when $x=0.$

Answer 1

Let $I=log\left(\frac{dy}{dx}\right)=3x+4y$

$\Rightarrow \frac{dy}{dx}=e^{3x}.e^{4y}$

$\iff e^{-4y}dy=e^{3x}dx$

$\iff \int e^{-4y}dy=\int e^{3x}dx\iff \frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}+C$

Putting $y=0,x=0$

$C=\frac{-7}{12}$

Therefore, the required solution is $\frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}-\frac{7}{12}$

i.e., $4e^{3x}+3e^{-4y}=7$