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Giving and using the formula for a solution of an ordinary differential equation of order $n$, calculate the exact solution of the differential equation $$y''' - 3 y' + 2 y = 9 e^x , x > 0$$

I know that the solution $y$ is $y(x) = y_{part} + y_{hom}$ but how we can compute $y_{part}$? and I know that $y_{hom}= c_1 y_1 + c_2 y_2 + c_3 y_3$ but how can we compute $y_i$, $c_i$?

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We are given:

$$\tag 1 y''' - 3 y' + 2 y = 9 e^x , x > 0$$


To find the homogeneous solution, we have the characteristic polynomial as:

$$m^3 - 3m + 2 = 0 \rightarrow m_1 = -2, m_2 = 1, m_3 = 1$$

As you can see, we have a single root, $-2$ and a double root, $1$.

How do you write the homogeneous solution, $y_h$, from these roots?

For the particular solution, $y_p$, compare what you got for $y_h$ with the RHS of $(1)$. What do you notice? Using those observations, guess at a particular solution, then solve for the constants by plugging that guess back into $(1)$.

Your solution will be:

$$y(x) = y_c(x) + y_p(x) = c_1e^{-2 x}+c_2 e^x+c_3 e^x x + \frac{3 e^x x^2}{2}$$

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Nice approach and write-up >+< – amWhy Jun 30 '13 at 0:04

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