If y=a e^ 2 x +b e^ -x , then show that d^2 y d x^2 -d y d x -2 y=0.

y=a e^ 2 x +b e^ -x (1) d y d x =2 a e^ 2 x -b e^ -x (2) d^2 y d x^2 =4 a e^ 2 x +b e^ -x (3) Putting the values on LHS l =d^2 y d x^2 -d y d x -2 y = (4 a e^ 2 x +b e^ -x )- (2 a e^ 2 x -b e^ -x )-2 (a e^ 2 x +b e^ -x ) =4 a e^ 2 x +b e^ -x -2 a e^ 2 x +b e^ -x -2 a e^ 2 x -2 b e^ -x =0

If y=a e^ 2 x +b e^ -x , then show that d^2 y d x^2 -d y d x -2 y…

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Question

If $y=a e^{2 x}+b e^{-x}$, then show that $\frac{d^2 y}{d x^2}-\frac{d y}{d x}-2 y=0$.

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Answer

$y=a e^{2 x}+b e^{-x}\quad \ldots(1)$
$\frac{d y}{d x}=2 a e^{2 x}-b e^{-x}\quad \ldots(2)$
$\frac{d^2 y}{d x^2}=4 a e^{2 x}+b e^{-x}\quad \ldots(3)$
Putting the values on LHS
$
\begin{array}{l}
=\frac{d^2 y}{d x^2}-\frac{d y}{d x}-2 y \\
=\left(4 a e^{2 x}+b e^{-x}\right)-\left(2 a e^{2 x}-b e^{-x}\right)-2\left(a e^{2 x}+b e^{-x}\right) \\
=4 a e^{2 x}+b e^{-x}-2 a e^{2 x}+b e^{-x}-2 a e^{2 x}-2 b e^{-x} \\
=0
\end{array}
$

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