Show that y=ae^ 2x +be^ -x is a solution of the differential equation d^2y dx^2 - dy dx -2y=0

We have, y=ae^ 2x +be^ -x ...(1) Differentiating both sides of equation (1) with respect to 3, we get dy dx =ae^ 2x +be^ -x ...(2) Differentiating both sides of equation (2) with respect to 3, we get d^2y dx^2 =4ae^ 2x +be^ -x d^2y dx^2 =2ae^ 2x -be^ -x +2ae^ 2x +2be^ -x d^2y dx^2 = (2ae^ 2x -be^ -x )+2 (ae^ 2x +2be^ -x ) d^2y dx^2 = dy dx +2y d^2y dx^2 - dy dx -2y=0 Hence, the given function is the solution to the given differential equation.

Show that y=ae^ 2x +be^ -x is a solution of the differential equa…

Download App

Question

Show that $\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$

✓

Answer

We have,

$\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}\ ...(1)$

Differentiating both sides of equation (1) with respect to 3, we get

$\frac{\text{dy}}{\text{dx}}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}\ ...(2)$

Differentiating both sides of equation (2) with respect to 3, we get

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=4\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{ae}^{2\text{x}}-\text{be}^{-\text{x}}+2\text{ae}^{2\text{x}}+2\text{be}^{-\text{x}}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\big(2\text{ae}^{2\text{x}}-\text{be}^{-\text{x}}\big)+2\big(\text{ae}^{2\text{x}}+2\text{be}^{-\text{x}}\big)$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}+2\text{y}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$

Hence, the given function is the solution to the given differential equation.