Show that the function $y = e^{3 x} (A + B x)$ is a solution of the differential equation $\frac{d^{2} y}{d x^{2}} - 6 \frac{d y}{d x} + 9 y = 0$ .

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Solution

given differential equation is $\frac{d^{2} y}{d x^{2}} - 6 \frac{d y}{d x} + 9 y = 0.... (1)$
$y = e^{3 x} (A + B x)$
diiferetiating
$= 3 A e^{3 x} + B (3 x + 1) e^{3 x}$
differentiating again
$= 9 A e^{3 x} + 3 B (3 x + 1) e^{3 x} + 3 B e^{3 x}$
Substituting in (1)
$9 A e^{3 x} + 9 B x e^{3 x} + 6 B e^{3 x} - 18 A e^{3 x} - 18 B x e^{3 x} - 6 B e^{3 x} + 9 A e^{3 x} + 9 B x e^{3 x} = 0$
$0 = 0$
$∴$ $y = e^{3 x} (A + B x)$ is a solution of (1)

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