# If  y=ce^(2x)+De^(-2x)  then show that  y'' -4y = 0 ?

Aug 15, 2017

$y ' ' - 4 y = 0$

#### Explanation:

This is an inverse problem. Given a solution, find a candidate differential equation which has is as solution.

Assuming the differential equation is second-order linear homogeneous such as

${c}_{1} y ' ' + {c}_{2} y ' + {c}_{3} y = 0$

after substituting the solution we have

$\left(4 {c}_{1} - 2 {c}_{2} + {c}_{3}\right) {D}_{0} {e}^{- 2 x} + {C}_{0} \left(4 {c}_{1} + 2 {c}_{2} + {c}_{3}\right) {e}^{2 x} = 0$

This relationship must be true for all $x$ so

$\left\{\begin{matrix}4 {c}_{1} - 2 {c}_{2} + {c}_{3} = 0 \\ 4 {c}_{1} + 2 {c}_{2} + {c}_{3} = 0\end{matrix}\right.$

now solving for ${c}_{1} , {c}_{2}$ we obtain

${c}_{1} = - {c}_{3} / 4 , {c}_{2} = 0$ so the differential equation is

$- {c}_{3} / 4 y ' ' + {c}_{3} y = 0$ or

$y ' ' - 4 y = 0$

Aug 15, 2017

Refer to the Explanation.

#### Explanation:

We note that, the given eqn. $y = C {e}^{2 x} + D {e}^{- 2 x} \ldots \ldots \left(1\right) ,$ contains

2 arbitrary constants.

Therefore, the reqd. Diff. Eqn. must be of Second Order.

To find it, we diif. $\left(1\right)$ twice.

$y = C {e}^{2 x} + D {e}^{- 2 x} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \left(C {e}^{2 x}\right) 2 + \left(D {e}^{- 2 x}\right) \left(- 2\right) , \mathmr{and} ,$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left\{C {e}^{2 x} - D {e}^{- 2 x}\right\} ,$

$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left[2 \left\{C {e}^{2 x} - D {e}^{- 2 x}\right\}\right] ,$

$= 2 \left[\left(C {e}^{2 x}\right) \left(2\right) - \left(D {e}^{- 2 x}\right) \left(- 2\right)\right] ,$

$\Rightarrow \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 4 \left[C {e}^{2 x} + D {e}^{- 2 x}\right] = 4 y ,$

$\Rightarrow \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 - 4 y = 0 ,$ is the Desired Diff. Eqn.

Aug 15, 2017

$y ' ' - 4 y = 0$

#### Explanation:

We have:

$y = c {e}^{2 x} + D {e}^{- 2 x}$ .... [A}

As others have indicated we are not solving a second order Differentiation Equation with constant coefficients, but rather forming one given the solution.

Recognizing the solution is that of a second order Differentiation Equation with constant coefficients we can instantly write down the appropriate DE.

The Auxiliary Equation that produced this solution would require two distinct real solution, $m = 2$ and m=-2)

Hence the associated Auxiliary Equation would be:

$\left(m - 2\right) \left(m + 2\right) = 0 \implies {m}^{2} - 4 = 0$

Hence the DE associated with this Auxiliary Equation is:

$y ' ' + 0 y ' - 4 y = 0$
$y ' ' - 4 y = 0$