Question

Solve the Differential Equation `y'' - 4 y' + 4y = 2 e^(2x)`?

Answer 1

`y_g = e^(2 x) ( x^2 + 2 x + 1 )`

Explanation:

Method of Undetermined Coefficients

Start with the homogeneous equation and the complementary solution :

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

This has characteristic equation:

`lambda^2 - 4lambda + 4 = 0 implies (lambda - 2)^2 = 0`

Repeated roots mean that, in lieu of the usual solution `y_c = alpha e^(lambda_1 x) + beta e^(lambda_2 x)`, we look here for a solution in the form:

`y_c = e^(2 x) ( alpha x + beta )`

And because the non-homogeneous equation already has a `e^(2x)` term, we must look at a particular solution in the form:

`y_p = gamma x^2 e^(2x)`
`implies y' = 2 gamma x e^(2x) + 2 gamma x^2 e^(2x)`
`implies y'' = 2 gamma e^(2x) + 8 gamma x e^(2x) + 4 gamma x^2 e^(2x)`

Putting these into the equation:

`2 gamma e^(2x) + 8 gamma x e^(2x) + 4 gamma x^2 e^(2x) - 4 (2 gamma x e^(2x) + 2 gamma x^2 e^(2x)) + 4 gamma x^2 e^(2x) = 2 e^(2x)`

`implies gamma = 1` and `y_p = x^2 e^(2x)`

The general solution is: `y_g = y_c + y_p`

`y_g = e^(2 x) ( alpha x + beta ) + x^2 e^(2x)`

`= e^(2 x) (x^2 + alpha x + beta )`

Now applying the IV's:

`y(0) = 1 implies beta = 1 implies y_g = e^(2 x) (x^2 + alpha x + 1 )`

`y' = 2 e^(2 x) (x^2 + alpha x + 1 ) + e^(2 x) (2 x + alpha )`

`= e^(2 x) (2x^2 + (2 alpha + 2) x + (2 + alpha) )`

And from the second IV, `y'(0) = 4 implies alpha = 2`

`y_g = e^(2 x) ( x^2 + 2 x + 1 )`

Answer 2

The solution is `y(x)=(x+1)^2e^(2x)`

Explanation:

`y''-4y'+4y=2e^(2x)`

This is a second order linear, non-homogeneous differential equation.

The general solution can be written as

`y=y_p+y_h`

`y_h` is the solution to `y''-4y'+4y=0`

`y_p` is the particular solution

The caracteristic equation is

`r^2-4r+4=0`

`(r-2)^2=0`

We have a double root

The solution without the LHS is

`y_h=(Ax+B)e^(2x)`

For the particular solution, we try

`y_p=Cx^2e^(2x)`

`y'=C(2xe^(2x)+2x^2e^(2x))`

`y'=2Ce^(2x)(x+x^2)`

`y''=2C(2e^(2x)(x+x^2)+e^(2x)(1+2x))`

Putting these in the equation

`2C(2e^(2x)(x+x^2)+e^(2x)(1+2x))-4*2Ce^(2x)(x+x^2)+4*Cx^2e^(2x)=2e^(2x)`

`4Cx+4cx^2+2C+4Cx-8Cx-8Cx^2+4cx^2=2`

`C=1`

So, the particular solution is

`y_p=x^2e^(2x)`

The general solution is

`y=(Ax+B)e^(2x)+x^2e^(2x)`

`y(0)=B=1`

`y'=Ae^(2x)+2(Ax+B)e^(2x)+2xe^(2x)+2x^2e^(2x)`

`y'(0)=A+2B=4`

`A=4-2=2`

Finally, we have

`y(x)=(2x+1)e^(2x)+x^2e^(2x)`

`=e^(2x)(x^2+2x+1)`

`=(x+1)^2e^(2x)`

Answer 3

The two other solutions quite clearly demonstrate how to solve the complementary solution of the homogeneous equation.

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

As this is fairly standard text book stuff which solves the Auxiliary equation to form a guaranteed solution based of the roots of the equation

The interesting part is find the solution of the particular function, and where did the magic "lets try" `y=Cx^2e^(2x)` come from?

Basically it is down to practice & experience but there is a solid method to find the particular solution using the Wronskian. It does, however, involve a lot more work:

Having established that the solution of the homogeneous equation is (see other answers):

`y_c = (Ax+B)e^(2x)`
`\ \ \ = Axe^(2x) + Be^(2x)`

The reason this form of solution works is that the two individual components `xe^(2x)` and `e^(2x)` are linearly independent.

Once we have two linearly independent solutions say `y_1(x)` and `y_2(x)` then the particular solution of the general DE;

`ay'' +by' + cy = p(x)`

is given by:

`y_p = v_1y_1 + v_2y_2 \ \ `, which are all functions of `x`

Where:

`v_1 = -int \ (p(x)y_2)/(W[y_1,y_2]) \ dx`
`v_2 = \ \ \ \ \ int \ (p(x)y_1)/(W[y_1,y_2]) \ dx`

And, `W[y_1,y_2]` is the wronskian; defined by the following determinant:

`W[y_1,y_2] = | ( y_1,y_2), (y'_1,y'_2) |`

So for our equation:

`p(x) = 2e^(2x)`
`y_1 \ \ \ = xe^(2x) => y'_1 = 2xe^(2x) + e^(2x)`
`y_2 \ \ \ = e^(2x) \ \ => y'_2 = 2e^(2x)`

So the wronskian for this equation is:

`W[y_1,y_2] = | ( xe^(2x),,e^(2x)), (2xe^(2x) + e^(2x),,2e^(2x)) |`
`" " = (xe^(2x))(2e^(2x)) -(e^(2x))(2xe^(2x) + e^(2x))`
`" " = (e^(2x))(2xe^(2x) -2xe^(2x) - e^(2x))`
`" " = -e^(4x)`

So we form the two particular solution function:

`v_1 = -int \ (p(x)y_2)/(W[y_1,y_2]) \ dx`
`\ \ \ = -int \ ((2e^(2x))(e^(2x)))/((-e^(4x))) \ dx`
`\ \ \ = int \ 2 \ dx`
`\ \ \ = 2x`

And;

`v_2 = \ \ \ \ \ int \ (p(x)y_1)/(W[y_1,y_2]) \ dx`
`\ \ \ = int \ ( (2e^(2x))(xe^(2x)) ) / (-e^(4x)) \ dx`
`\ \ \ = int \ -2x \ dx`
`\ \ \ = -x^2`

And so we form the Particular solution:

`y_p = v_1y_1 + v_2y_2`
`\ \ \ = (2x)(xe^(2x)) + (-x^2)(e^(2x))`
`\ \ \ = e^(2x)(2x^2-x^2)`
`\ \ \ = x^2e^(2x)`

Which is the same particular solution as the other answers produced, leading to the general solution:

`y(x) = y_c + y_p`
`\ \ \ \ \ \ \ = Axe^(2x) + Be^(2x) + x^2e^(2x)`

Which then leads to the same specific solution of the other answers.

Answer 4

Another approach using the fact that `d/dx` is a linear operator.

Explanation:

Mary Boas teaches this in a way I not see very widely used, but which also obviates the need for the experienced guesswork.

If we note that `D = d/dx` is a linear operator, we can take the original equation and write it as this:

`y'' - 4 y' + 4y = 2 e^(2x)`

`implies D^2y - 4 D y + 4y = 2 e^(2x)`

`(D- 2)(D-2)y = 2 e^(2x)`

So it already looks like the eigenvalue form you'd get from a linear system, but is non-homogeneous. Then we can say that `z(x) = (D-2)y(x)` so we have this more trivial Integrating Factor problem:

`(D- 2)z = 2 e^(2x)`

`z'- 2z = 2 e^(2x)`

Integrating factor `=exp ( int -2 \ dx)`

`e^(-2x)(z'- 2z) = 2 e^(2x) e^(-2x)`

`(z e^(-2x))' = 2`

`z e^(-2x) = 2 x + alpha`

`(D-2)y = e^(2x) ( 2 x + alpha)`

`y'-2y = e^(2x) ( 2 x + alpha)`

(You can even apply the `y'(0) = 4` I.V. here (!!).)

So, it's another Integrating Factor: `=exp ( int -2 \ dx)`

`e^(-2x) (y'-2y) = 2 x + alpha`

`(e^(-2x) y)' = 2 x + alpha`

`e^(-2x) y = x^2 + alpha x + beta`

`y = e^(-2x)( x^2 + alpha x + beta )`

And then apply the IV's as before.

Boas is my personal favourite maths book, but it's for scientists, not mathematicians. But this is a real good way to go at repeated eigenvalues, which are tricky.

Answer 5

`y = e^(2x) (x+1)^2`

Explanation:

Laplace Transform

`y'' - 4 y' + 4y = 2 e^(2x)`

`p^2 Y - p y_o - y_0' - 4 (pY - y_o) + 4Y = 2 /(p-2)`

We have the IV's at `x = 0`, so:

`p^2 Y - p - 4 - 4 pY + 4 + 4Y = 2 /(p-2)`

`Y (p^2 - 4 p + 4) = p + 2 /(p-2)`

`Y = p/(p -2)^2 + 2 /(p-2)^3`

From Standard Tables, here using Mary Boas 3rd Ed, L6 and L18, pp469-470:

`y = e^(2x) (1 + 2x) + e^(2x) x^2`

`= e^(2x) (x+1)^2`