Question

The particular solution of the differential equation $log(dy/dx)=3x+4y$ at $x=0$ and $y=0$ is

• A. $4e^{3x}-3e^{-4y}=7$
• B. $4e^{3x}+3e^{-4y}=0$
• C. $4e^{3x}+3e^{-4y}=7$
• D. $4e^{3x}-3e^{-4y}=0$

Answer 1

Answer: (C)

The given equation is $log(\frac{dy}{dx})=3x+4y$
$\Rightarrow \frac{dy}{dx}=e^{3x+4y}$
$\Rightarrow \frac{dy}{dx}=e^{3x}\cdot e^{4y}$
$\Rightarrow e^{-4y}\cdot dy=e^{3x}\cdot dx$
Integrating both sides, we get
$\int e^{-4y}\cdot dy=\int e^{3x}\cdot dx$
$\Rightarrow \frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}+C$
$\Rightarrow \frac{e^{-4y}}{4}+\frac{e^{3x}}{3}+C=0$
$\Rightarrow 3e^{-4y}+4e^{3x}+12C=0...(i)$
Now it is given that when $x=0,y=0$.
$\therefore 3+4+12C=0$
$\Rightarrow C=-\frac{7}{12}$
$\therefore$ From $(i),4e^{3x}+3e^{-4y}+12(-\frac{7}{12})=0$
$\Rightarrow 4e^{3x}+3e^{-4y}=7$,
which is the required solution.