Question

Solve the given differential equation by undetermined coefficients. 1 4 y'' + y' + y = x2 − 2x

Answer 1

Answer:
`y_(gen) = c_(1) e^{(x)/(8) }cos((√(15))/(8)x) + c_(2) e^{(x)/(8)}sin((√(15))/(8) x) + x^(2) -4x - 4`

Explanation:

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

First, we find the homogenous solution to this differential equation by solving
`4y^(2) + y + 1 = 0` and we will get
`y = (-1 +/- i√(15))/(8)`.

Recall that the homogenous solution, when roots are in complex form (
`\alpha +/-\beta i`), is
`y_(hom) = c_(1) e^(\alpha x)cos(\beta x) + c_(2) e^(\alpha x)sin(\beta x)` when
`c_(1)` and
`c_(2)` are undetermined coefficients. So substituting
`\alpha = -1/8` and
`\beta = (√(15) )/(8)`, we would get the equation
`y_(hom) = c_(1) e^{(x)/(8) }cos((√(15))/(8)x) + c_(2) e^{(x)/(8)}sin((√(15))/(8) x)`

However, we are not done. We still have to find the particular solution to this inhomogenous equation by using undetermined coeffeicients. This means we are going to take the term g(x), or the term containing no y terms, and find a match to substitute it.

In this case, it is a polynomial, so the substitute must be a polynomial of the same degree. We'll say it is
`Ax^(2) + Bx + C`. If
`y_(par) = Ax^(2) + Bx + C`, that means
`y' = 2Ax + B` and
`y'' = 2A`. Substitute the three into the differential equation, we get

`4(2A) + 2Ax + B + Ax^(2) +Bx + C = x^(2) - 2x`

From here we can solve for
`A, B, C` which is 1, -4, -4. And we can just substitute that into out particular solution to get
`y_(par) = x^(2) -4x - 4`. Combining the homogenous solution and the particular solution, we get that the general solution is

`y_(gen) = c_(1) e^{(x)/(8) }cos((√(15))/(8)x) + c_(2) e^{(x)/(8)}sin((√(15))/(8) x)+ x^(2) -4x - 4`