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Use the method of reduction of order to find a solution of the given nonhomogeneous equation.?
The indicated y1(x) is a solution of the associated HOMOGENEOUS equation. Determine a second solution of the homogeneous equation and a particular soution of the nonhomogeneous equation.
(meaning I need a solution y=c1y1 + c2y2 + yp)
y'' - 4y' +3y = x ; y1(x) = e^x
So far, I found the second solution of the homogeneous equation to be 1/2e^(3x), however I have no idea where to go from here.
My book says if the g(x) is in the form of a linear function of x, then yp is in the form Ax+B, but I have no idea how to find it... Can anyone help?
2 Answers
- kbLv 78 years agoFavorite Answer
For reduction of order, we assume that y = e^x * v.
Differentiating:
y' = e^x (v' + v)
y'' = e^x (v'' + 2v' + v).
Substituting this into the original DE:
e^x (v'' + 2v' + v) - 4e^x (v' + v) + 3e^x * v = x
==> v'' - 2v' = xe^(-x), a first order linear DE in y'
Multiply both sides by the integrating factor e^(-2x):
(d/dx)(e^(-2x) v') = xe^(-3x)
Integrate both sides:
e^(-2x) v' = xe^(-3x)/(-3) - e^(-3x)/9 + 2A
==> v' = (-1/3) xe^(-x) - (1/9)e^(-x) + 2Ae^(2x)
Integrate both sides again:
v = (-1/3)(-xe^(-x) - e^(-x)) + (1/9)e^(-x) + Ae^(2x) + B
...= (1/9)(3xe^(-x) + 4e^(-x)) + Ae^(2x) + B.
Since y = e^x * v, we obtain
y = (1/9)(3x + 4) + Ae^(3x) + Be^x.
I hope this helps!