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A particular solution of the initial value differential equation $\log \frac{d y}{d x} = 7 x + 8 y, y (0) = 0$ is

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$15 y = 8 e^{- 8 y} + 7 e^{- 7 y}$

$15 x = 7 e^{- 8 y} + 8 e^{7 x}$

$8 e^{- 8 y} + 7 e^{7 x} = 15$

$7 e^{- 8 y} + 8 e^{7 x} = 15$

answer is D.

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Detailed Solution

$\frac{d y}{d x} = e^{7 x + 8 y} = e^{7 x} e^{8 y} \Rightarrow e^{- 8 y} d y = e^{7 x} d x$

so $\frac{e^{- 8 y}}{- 8} = \frac{e^{7 x}}{7} + C$ Since $y (0) = 0$ so $- \frac{1}{8} = \frac{1}{7} + C$

$\Rightarrow C = - \frac{1}{8} - \frac{1}{7} = - \frac{15}{56} .$ Thus $\frac{e^{- 8 y}}{- 8} = \frac{e^{7 x}}{7} - \frac{15}{56}$

$\Rightarrow - 7 e^{- 8 y} = 8 e^{7 x} - 15 \Rightarrow 7 e^{- 8 y} + 8 e^{7 x} = 15$

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