The D.E whose solution is $y = C_{1} e^{3 x} + C_{2} e^{5 x}$ is:

y_{2} + 2 y_{1} + 15 y = 0

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y_{2} + 8 y_{1} + 15 y = 0

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y_{2} + 8 y_{1} - 15 y = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y_{2} - 8 y_{1} + 15 y = 0

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Solution

The correct option is D $y_{2} - 8 y_{1} + 15 y = 0$
$y = C_{1} e^{3 x} + C_{2} e^{5 x}$
$\Rightarrow y_{1} = 3 C_{1} e^{3 x} + 5 C_{2} e^{5 x}$ ----------(1)
$y_{2} = 9 C_{1} e^{3 x} + 25 C_{2} e^{5 x}$ ------------(2)
$(2) - 3 \times (1)$
$\Rightarrow y_{2} - 3 y_{1} = 10 C_{2} C^{5 x}$
$\Rightarrow C_{2} e^{5 x} = \frac{y_{2} - 3 y_{1}}{10}$
$\Rightarrow y_{1} = 3 C_{1} e^{3 x} + \frac{(y_{2} - 3 y_{1})}{2}$
$\frac{2 y_{1} - (y_{2} - 3 y_{1})}{6} - C_{1} e^{3 x} = \frac{5 y_{1} - y_{2}}{6}$
$\Rightarrow y = \frac{5 y_{1} - y_{2}}{6} + \frac{y_{2} - 3 y_{1}}{10}$
$\Rightarrow 2 y = 25 y_{1} - 5 y_{2} + 3 y_{2} - 9 y_{1}$
$15 \times 2 y = 16 y_{1} - 2 y_{2}$
$\Rightarrow y_{2} - 8 y_{1} + 15 y = 0$

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