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Particular Solution of a Differential Equation

Find a partic...

Question

Find a particular solution for the following differential equation.
$y^{'} - 4 y^{'} - 12 y = t e^{4 t}$

y (t) = \frac{1}{32} (3 t + 1) e^{4 t}

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y (t) = - \frac{1}{18} (3 t + 1) e^{4 t}

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y (t) = - \frac{1}{36} (3 t + 1) e^{4 t}

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None of these

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Solution

The correct option is D None of these
Given, $y^{'} - 4 y^{'} - 12 y = t e^{4 t}$
$\Rightarrow - {3 y}^{'} - 12 y = t e^{4 t} \Rightarrow y^{'} + 4 y = - \frac{t e^{4 t}}{3} \Rightarrow \frac{d y}{d t} + 4 y = - \frac{1}{3} t e^{4 t}$
This is a linear form of differential equation with
$I.F. = e^{\int 4 d t} = e^{4 t}$
Solution of the differential equation is
$y (I.F.) = \int Q (I.F.) d t y e^{4 t} = \int \frac{- t e^{4 t}}{3} e^{4 t} d t y e^{4 t} = \frac{- 1}{3} \int t e^{8 t} d t y e^{4 t} = \frac{- 1}{3} {t \int e^{8 t} - \int \frac{d}{d t} (t) \int e^{8 t}} y e^{4 t} = \frac{- 1}{3} {\frac{t e^{8 t}}{8} - \int \frac{e^{8 t}}{8}} y e^{4 t} = \frac{- 1}{3} {\frac{t e^{8 t}}{8} - \frac{e^{8 t}}{64}} y e^{4 t} = \frac{- e^{8 t}}{3} {\frac{8 t - 1}{64}} y = \frac{- e^{4 t}}{192} (8 t - 1)$
So, option D is correct.

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