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If both of these are part of the homogenous solution, we should try:
We should not try higher powers of x, as we would need at least 3 derivatives of g ( x) ke rx
y p Ax 3e rx to get a to appear. The goal is to ask: what could I g ( x) ke rx
derive twice, once, and none, combine them together, and make appear somehow (this is a tough question). Example:
y ' '2 y ' y 2e x Say we had: Then we will first find the homogenous system:
Auxiliary Equation is: r2‐2r+1=0 Factoring/Quadratic Formula gives: r=1
This means our solution to the homogenous system is: yh c1e x c2 xe x yp e
y p Axe
Since is part of our homogenous solution, would try , but it is also part of the homogenous solution. This means we are forced to try:
x x y p Ax 2 e x
The goal will be to find out which A will make yp a solution to the ODE. Example Continued:
y ' '2 y ' y 2e x Sub in yp into: ( Ax 2 e x )' '2( Ax 2 e x )' Ax 2 e x 2e x
(2 Axe x Ax 2 e x )'2(2 Axe x Ax 2 e x ) Ax 2 e x 2e x
2 Ae x 2 Axe x 2 Axe x Ax 2 e x 4 Axe x 2 Ax 2 e x Ax 2 e x 2e x
2 Ae x 2e x
Here we match up the coefficients of the terms. This means 2A=2. We can easily solve this to get A=1.
This means our y p x 2e x This makes our final solution: y = yh+yp = c1e x c2 xe x x 2 e x Strategy for Undetermined Coefficients
Step 1: Solve the homogenous solution yh for the ODE (ie when g(x)=0)
Step 2: Determine the candidate for the particular solution yp (with coefficients that are not known, so use A, B, C, … for the coefficients). This solution must not be part of the homogenous solution (as the homoge...
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- Summer '00
- Calculus, SmartPen Lectures, Lecture Notes, MATH1004, Kyle Harvey