Solve: $y^{'} + 3 y = r^{3 x} y^{2}$

y = b \frac{1}{(c + x) r^{- 3 x}}

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y = b \frac{1}{(c - x) r^{3 x}}

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y = b \frac{1}{(c - x) r^{2 x}}

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y = b \frac{1}{(c + x) r^{- 2 x}}

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Solution

The correct option is B $y = b \frac{1}{(c - x) r^{3 x}}$
$3 y + \frac{d y}{d x} = r^{3 x} y^{2} \Rightarrow - \frac{\frac{d y}{d x}}{y^{2}} - \frac{3}{y} = - r^{3 x}$
Let $v = \frac{1}{y} \Rightarrow \frac{d v}{d x} = - \frac{\frac{d y}{d x}}{y^{2}}$
$\frac{d v}{d x} - 3 v = - r^{3 x}$
Let $μ = e^{\int - 3 d x} = e^{- 3 x}$
$\frac{\frac{d v}{d x}}{e^{3 x}} - \frac{3 v}{e^{3 x}} = - \frac{r^{3 x}}{e^{3 x}}$
Substitute $- 3 e^{- 3 x} = \frac{d}{d x} (e^{- 3 x})$
$\frac{\frac{d v}{d x}}{e^{3 x}} + \frac{d}{d x} (e^{- 3 x}) v = - \frac{r^{3 x}}{e^{3 x}}$

Using $g \frac{d f}{d x} + f \frac{d g}{d x} = \frac{d}{d x} (f g)$
$\frac{d}{d x} (\frac{v}{e^{3 x}}) = - \frac{r^{3 x}}{e^{3 x}} \Rightarrow \int \frac{d}{d x} (\frac{v}{e^{3 x}}) d x = \int - \frac{r^{3 x}}{e^{3 x}} d x \Rightarrow \frac{v}{e^{3 x}} = - \frac{r^{3 x}}{3 (log r - 1) e^{3 x}}$
Hence $y = b \frac{1}{(c - x) r^{3 x}}$

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