Problem:
y'' + y' = x
Solution:
Again, we solve the homogeneous equation $ z'' + z' = 0 $, the solution is $ z = Ae^{-x} + B $.
Next we guess the particular solution to $ w'' + w' = x $ is a quadratic polynomial $ w(x) = Px^2 + Qx + R $, the derivatives are
$ w'(x) = 2Px + Q $
$ w''(x) = 2P $.
Therefore we have $ 2P + 2Px + Q = x $, so we get $ P = \frac{1}{2} $ and $ Q = -1 $.
The final answer is therefore $ y = Ae^{-x} + \frac{1}{2}x^2 - x + B $.
y'' + y' = x
Solution:
Again, we solve the homogeneous equation $ z'' + z' = 0 $, the solution is $ z = Ae^{-x} + B $.
Next we guess the particular solution to $ w'' + w' = x $ is a quadratic polynomial $ w(x) = Px^2 + Qx + R $, the derivatives are
$ w'(x) = 2Px + Q $
$ w''(x) = 2P $.
Therefore we have $ 2P + 2Px + Q = x $, so we get $ P = \frac{1}{2} $ and $ Q = -1 $.
The final answer is therefore $ y = Ae^{-x} + \frac{1}{2}x^2 - x + B $.
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