Question

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

Answer 1

`y''-4y=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

`(4 c_1 - 2 c_2 + c_3) D_0 e^(-2 x) + C_0 (4 c_1 + 2 c_2 + c_3) e^(2 x)=0`

This relationship must be true for all `x` so

`{(4 c_1 - 2 c_2 + c_3=0),(4 c_1 + 2 c_2 + c_3=0):}`

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''-4y=0`

Answer 2

Refer to the Explanation.

Explanation:

We note that, the given eqn. `y=Ce^(2x)+De^(-2x)......(1),` contains

2 arbitrary constants.

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

To find it, we diif. `(1)` twice.

`y=Ce^(2x)+De^(-2x) rArr dy/dx=(Ce^(2x))2+(De^(-2x))(-2), or,`

`dy/dx=2{Ce^(2x)-De^(-2x)},`

`:. (d^2y)/dx^2=d/dx[2{Ce^(2x)-De^(-2x)}],`

`=2[(Ce^(2x))(2)-(De^(-2x))(-2)],`

`rArr (d^2y)/dx^2=4[Ce^(2x)+De^(-2x)]=4y,`

`rArr (d^2y)/dx^2-4y=0,` is the Desired Diff. Eqn.

Answer 3

`y'' -4y = 0`

Explanation:

We have:

`y=ce^(2x)+De^(-2x)` .... [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:

`(m-2)(m+2) = 0 => m^2-4 = 0`

Hence the DE associated with this Auxiliary Equation is:

`y'' + 0y'-4y = 0`
`y'' -4y = 0`