Differential equations, Linear Equations: Find value of B so that a function is a solution to homogeneous func.

Question

I need to find the value of B so the function $y=c_1e^{3x}+c_2e^{5x}+Be^{4x}$ is a solution to $y''-8y'+15y=6e^{4x}$.

Here's the work I have done. I know that $y_1=e^{3x}$ so I can use that to find some differentials: $y_1'=3e^{3x}$ and $y_1''=9e^{3x}$ substituting these values into the homogenous equation above just verifies that the functions are solutions to the nonhomogeneous equation (sorry if I have the homogenous and nonhomogeneous equations backward).

Anyway, after that verification, I would take $y_p=6e^{4x}$ and verify it too by subbing. However, this does not help me find the value of B.

I'm really not sure what to do with this question and would really appreciate some help. I've been looking at this for quite a while now.

Thank you.

Answer 1

$$y''-8y'+15y=6e^{4x}$$ Plug the particular solution in the equation $$16Be^{4x}-32Be^{4x}+15Be^{4x}=6e^{4x}$$ $$\implies (B+6)e^{4x}=0 \implies B=-6$$

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Solving the homogeneous equation you get $$y''-8y'+15y=6e^{4x}$$ $$r^2-8r+15 =0 \implies (r-3)(r-5)=0$$ $$y(x)=c_1e^{3x}+c_2e^{5x}$$ $y_p=Be^{4x}$ is the particular solution of the inhomogeneous equation

Answer 2

Being myself very lazy, write the equation as $$y''-8y'+15y-6e^{4x}=0 \tag 1$$ Now, since you are given $$y=c_1e^{3x}+c_2e^{5x}+Be^{4x}\tag 2$$ $$y'=3c_1e^{3x}+5c_2e^{5x}+4Be^{4x}\tag 3$$ $$y''=9c_1e^{3x}+25c_2e^{5x}+16Be^{4x}\tag 4$$ Replace $(2)$, $(3)$ and $(4)$ in $(1)$ to get $$-(B+6) e^{4 x}=0$$