Question

Solution of $\log \left(\frac{dy}{dx}\right)=3x+4y,y\left(0\right)=0$ is

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

Answer 1

Answer: (D)

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

$\log \left(\frac{dy}{dx}\right)=3x+4yy\left(0\right)$

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

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

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

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

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

given $y\left(0\right)=0$

So ,$\frac{e^0}{-4}=\frac{e^0}{3}+c$

$\Rightarrow \frac{1}{-4}=\frac{1}{3}+c$

$\Rightarrow c=-\frac{1}{3}-\frac{1}{4}$

$\Rightarrow c=\frac{-7}{12}$

Therefore, solution of given D.E is

$\frac{-1}{4}{e}^{-4y}=\frac{e^{3x}}{3}-\frac{7}{12}$

$\Rightarrow -3e^{-4y}=4e^{3x}-7$

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