If $y = e^{2 x} (a x + b)$ , show that $y_{2} - 4 y_{1} + 4 y = 0$ .

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Solution

Given,

$y = e^{2 x} (a x + b)$

To prove: $y_{2} - 4 y_{1} + 4 y = 0$

Proof:

We have,

$y = e^{2 x} (a x + b)$ ...(i)

$y_{1} = \frac{d y}{d x} = a e^{2 x} + 2 e^{2 x} (a x + b) . . . (ii) y_{2} = 2 a \times e^{2 x} + 4 e^{2 x} (a x + b) + 2 a e^{2 x} = 4 a e^{2 x} + 4 e^{2 x} (a x + b) . . . (iii) LHS = y_{2} - 4 y_{1} + 4 y = 4 a e^{2 x} + 4 e^{2 x} (a x + b) - 4 a e^{2 x} - 8 e^{2 x} (a x + b) + 4 e^{2 x} (a x + b) = 0 = RHS$

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