ODE $ y''-3y'+2y=4(2-a)e^{-ax}$

Question

I have to solve the system

$y'=-y-2z+2e^{-ax}$,

$z'=3y+4z+e^{-ax}$.

I expressed z from the first equation and plugged it into the second equation. The result is equation $y''-3y'+2y=4(2-a)e^{-ax}$. How to solve this equation?

Any help is welcome. Thanks in advance.

Answer 1

$$y''-3y'+2y=4(2-a)e^{-ax}$$ Solve the homogeneous part first: $$y''-3y'+2y=0$$ Polynomial characteristic is $$r^2-3r+2=(r-1)(r-2)=0 $$ $$ \implies S_r=\{1,2\}$$ $$y_h=c_1e^x+c_2e^{2x}$$ For the particular solution it depends on $a$ if $a \ne -1,-2$ try the particular solution: $$y_p=Ae^{-ax}$$ $a$ if $a = -1$ or $a=-2$ try the particular solution: $$y_p=Axe^{-ax}$$ Plug this in the original inhomogeneous DE and find the constant A then the solution is $$y=y_h+y_p$$

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You can add both DE and solve a first order DE instead $$(y+z)'=2(y+z)+3e^{-ax}$$ $$((y+z)e^{-2x})'=3e^{-(a+2)x}$$ This is easy to solve. $$y=-z +3e^{2x}\int e^{-(a+2)x}dx$$ The integral depends on the value of a you have to discuss ($a \ne -2$ or $a=-2$). Plug y in this DE $$z'=3y+4z+e^{-ax}$$ Then you have a first DE in $z$ that's easy to integrate.

Answer 2

hint

To make it easy, put

$$y=Ye^{-ax}$$ $$y'=(Y'-aY)e^{-ax}$$ $$y''=(Y''-2aY'+a^2Y)e^{-ax}$$

Your last equation becomes

$$Y''-2aY'+a^2Y-3Y'+3aY+2Y=4(2-a)$$ or $$Y''-(2a+3)Y'+(a^2+3a+2)Y=4(2-a)$$

The discriminant is $$(2a+3)^2-4(a^2+3a+2)=1$$ A particular constant solution is $$Y_p=\frac{4(2-a)}{(a+1)(a+2)}$$ Thus $$Y=Ae^{(a+1)x}+Be^{(a+2)x}+Y_p$$ and $$y=Ae^x+Be^{2x}+\frac{4(2-a)}{(a+1)(a+2)}e^{-ax}$$ For $a\ne -1,-2$.