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Question
If \[y = e^{2x} \left( ax + b \right)\] show that \[y_2 - 4 y_1 + 4y = 0\] ?
Solution
Given, \[y = e^{2x} \left( ax + b \right)\] To prove: \[y_2 - 4 y_1 + 4y = 0\]
Proof:
We have,
\[y = e^{2x} \left( ax + b \right)\] ...(i)
\[y_1 = \frac{dy}{dx} = a e^{2x} + 2 e^{2x} (ax + b) . . . (ii)\]
\[ y_2 = 2a \times e^{2x} + 4 e^{2x} (ax + b) + 2a e^{2x} \]
\[ = 4a e^{2x} + 4 e^{2x} (ax + b) . . . (iii)\]
\[\text { LHS }= y_2 - 4 y_1 + 4y\]
\[ = 4a e^{2x} + 4 e^{2x} (ax + b) - 4a e^{2x} - 8 e^{2x} (ax + b) + 4 e^{2x} (ax + b)\]
\[ = 0 =\text { RHS}\]
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