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

[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 y_c;
y_c=c₁e^-4.2x+c₂e^-4.2xx

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

Any idea? Or I made some mistake?

Meric.

 

[HallsofIvy]

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.

 

[whatababy]

Oh I made a silly mistake in differentiate the Yp...

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