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