The solution of $y_{2} - 7 y_{1} + 12 y = 0$ is

y = C_{1} e^{3 x} + C_{2} e^{4 x}

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y = C_{1} x e^{3 x} + C_{2} e^{4 x}

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y = C_{1} e^{3 x} + C_{2} x e^{4 x}

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Solution

The correct option is A $y = C_{1} e^{3 x} + C_{2} e^{4 x}$

The given equation can be written as $(\frac{d}{d x} - 3) (\frac{d y}{d x} - 4 y) = 0 . . . . (i)$
If $\frac{d y}{d x} - 4 y = u$ then (I) reduces to $\frac{d u}{d x} - 3 u = 0$
$\Rightarrow \frac{d u}{u} = 3 d x \Rightarrow u = C_{1} e^{3 x}$ . Therefore, we have $\frac{d y}{d x} - 4 y = C_{1} e^{3 x}$ which is a linear equation whose I.F.is $e^{- 4 x}$ . So $\frac{d}{d x} (y e^{- 4 x}) = C_{1} e^{- x}$
$\Rightarrow y e^{- 4 x} = - C_{1} e^{- x} + C_{2} \Rightarrow y = C_{1} e^{3 x} + C_{2} e^{4 x}$

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