Question

If \[y = e^{2x} \left( ax + b \right)\]  show that  \[y_2 - 4 y_1 + 4y = 0\] ?

Answer 1

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)
\[y_1 = \frac{dy}{dx} = a e^{2x} + 2 e^{2x} (ax + b) . . . (ii)\]

\[ y_2 = 2a \times e^{2x} + 4 e^{2x} (ax + b) + 2a e^{2x} \]

\[ = 4a e^{2x} + 4 e^{2x} (ax + b) . . . (iii)\]

\[\text { LHS }= y_2 - 4 y_1 + 4y\]

\[ = 4a e^{2x} + 4 e^{2x} (ax + b) - 4a e^{2x} - 8 e^{2x} (ax + b) + 4 e^{2x} (ax + b)\]

\[ = 0 =\text {  RHS}\]