Find the sum of the digits of the value of $y (log 4)$ if $y_{2} - 7 y_{1} + 12 y = 0, y (0) = 2, y_{1} (0) = 7$ .

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Solution

The given equation can be written as
$(\frac{d y}{d x} - 3) (\frac{d y}{d x} - 4 y) = 0 (i)$
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 $\frac{d}{d x} (y e^{- 4 x}) = C_{1} e^{- x} \Rightarrow y e^{- 4 x} = - C_{1} e^{- x} + C_{2}$
$\Rightarrow y = C_{1} e^{3 x} + C_{2} e^{4 x}$ So
$2 = y (0) = C_{1} + C_{2} . y_{1} (0) = 3 C_{1} + 4 C_{2} = 7 \Rightarrow C_{1} = C_{2} = 1$ .
Hence $y = e^{3 x} + e^{4 x}$ . Thus $y (log 4) = 4^{3} + 4^{4} = 320$ .
Therefore the sum of the digits of the value of $y$ is $3 + 2 + 0 = 5$

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