# A tricky 2nd Oder ODE Problem,nonhomo delta=0 case

• whatababy
In summary, to solve the initial value problem y''+8.4y'+17.64y=e^{-4.2x}, with y(0)=1 and y'(0)=1, we first solve the corresponding homogeneous equation to get yc=c_1e^{-4.2x}+c_2xe^{-4.2x}. Then, we assume a particular solution yp=Axe^{-4.2x} and differentiate it to get yp'=2Ae^{-4.2x}-4.2Axe^{-4.2x} and yp''=2Ae^{-4.2x}-16.8Axe^{-4.2x}+17.64Axe^{-4.2

#### whatababy

solve the initial value problem:
y''+8.4y'+17.64y=e-4.2x
y(0)=1, y'(0)=1

y(x)=?

The way I tried to achieve is to solve the corresponding homo equation first:
y''+8.4y'+17.64y=0, which gives yc;
yc=c1e-4.2x+c2e-4.2xx

Then try to find yp, generally I would assume a yp=Ae-4.2x, but from the yc got above, clearly yp=Ae-4.2x or yp=Ae-4.2xx is not good. If I add one more 'x' in yp assumption,in which it seems A has to be 0, which is not right either...

Any idea? Or I made some mistake?

Meric.

Because both $e^{-4.2x}$ and $xe^{-4.2x}$ satisfy the homogeneous equation, you need "go up" one more x: try
$$y_p= x^2e^{-4.2x}$$

Was that what you meant by "add one more 'x' in yp assumption"?

If
$$y_p= Ax^2e^{-4.2x}$$
then
$$y_p'= 2Axe^{-4.2x}- 4.2Ax^2e^{-4.2x}$$
and
$$y_p"= 2Ae^{-4.2x}- 16.8Axe^{-4.2x}+ 17.64Ax^2e^{-4.2x}$$
Putting that into the differential equation, all the terms involving $xe^{-4.2x}$ or $x^2e^{-4.2x}$ will cancel leaving only $2Ae^{-4.2}= e^{-4.2x}$ so A is NOT 0.

Last edited by a moderator:
Oh I made a silly mistake in differentiate the Yp...

Yeah, you are right, I got the right answer now. Thanks