The solution of lndydx=3x+4y,y0=0 is

dydx=e3xe4y⇒e−4y−4=e3x3+C⇒C=−712

The solution of lndydx=3x+4y,y0=0 is

dydx=e3xe4y⇒e−4y−4=e3x3+C⇒C=−712

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The solution of $\ln (\frac{d y}{d x}) = 3 x + 4 y, y (0) = 0$ is

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

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

$\ln (\frac{d y}{d x}) = 3 x + 4 y, y (0) = 0$

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

answer is D.

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Detailed Solution

$\frac{d y}{d x} = e^{3 x} e^{4 y} \Rightarrow \frac{e^{- 4 y}}{- 4} = \frac{e^{3 x}}{3} + C \Rightarrow C = - \frac{7}{12}$

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The solution of $\ln (\frac{d y}{d x}) = 3 x + 4 y, y (0) = 0$ is