Solve y'''+2y''-5y'-6y = 100exp(-3x)+18exp(-x)

Question

For y'''+2y''-5y'-6y = 100exp(-3x) + 18exp(-x)

I got my yh = C1*exp(-3*x) + C2*exp(2*x) + C3*exp(-x)

and using modification rule, yp = x*C5*exp(-3*x) + x*C6*exp(-x)

solving for which I got yp = 10x*exp(-3*x) - 3x*exp(-x)

However the correct solution for yp is (7 + 10x)*exp(-3*x) +(0.5 - 3x)*exp(-x)

I don't quite understand where those two extra terms came from. Any insight into this will be highly appreciated. Thank you.

Answer 1

with $$y=e^{\lambda x}$$ we get $$\lambda^3+2\lambda^2-5\lambda-6=0$$ which is $$(\lambda-2)(\lambda+1)(\lambda+3)=0$$ so we have the complementary solution $$C_1e^{-3x}+C_2e^{-x}+C_3e^{2x}$$ for the particular solution make the ansatz$$y_p=a_1xe^{-3x}+a_2xe^{-x}$$

Answer 2

With $\textbf{D}$ operator: $$y'''+2y''-5y'-6y=100e^{-3x}+18e^{-x}$$ $$(\textbf{D}+3)(\textbf{D}+1)(\textbf{D}-2)y=100e^{-3x}+18e^{-x}$$ $$y=\frac{1}{(\textbf{D}+3)(\textbf{D}+1)(\textbf{D}-2)}(100e^{-3x}+18e^{-x})$$ then

with $u=\dfrac{1}{\textbf{D}-2}(100e^{-3x}+18e^{-x})$ we have $u'-2u=100e^{-3x}+18e^{-x}$ and $\boxed{u=-20e^{-3x}-6e^{-x}+C_1e^{2x}}$.

with $v=\dfrac{1}{\textbf{D}+1}(-20e^{-3x}-6e^{-x}+C_1e^{2x})$ we have $v'+v=-20e^{-3x}-6e^{-x}+C_1e^{2x}$ and $\boxed{v=10e^{-3x}-6xe^{-x}+\dfrac{C_1}{3}e^{2x}+C_2e^{-x}}$.

and with $y=\dfrac{1}{\textbf{D}+3}(10e^{-3x}-6xe^{-x}+\dfrac{C_1}{3}e^{2x}+C_2e^{-x})$ we have $y'+3y=10e^{-3x}-6xe^{-x}+\dfrac{C_1}{3}e^{2x}+C_2e^{-x}$ so $\boxed{y=(10x+C_3)e^{-3x}+(-3x+\dfrac12C_2)e^{-x}+\dfrac{C_1}{15}e^{2x}}$.