Question

The particular solution of differential equation $\log \left(\frac{dy}{dx}\right)=3x+4y$ is, when $y=0=x$

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

Answer 1

Answer: (C)

${4e}^{3x}+{3e}^{-4y}-7=0$

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

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

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

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

Integrating, we have

$\frac{e^{-4y}}{-4}-\frac{e^{3x}}{3}+c=0$

$\Rightarrow 3e^{-4y}+4e^{3x}-12c=0$

Since $x=0$ when $y=0$, substituting both the values, we have

$3+4-12c=0$

$\Rightarrow 12c=7$

The equation becomes $4e^{3x}+3e^{-4y}-7=0$