This problem is hindering my overall work

[Dimedrol]

This problem is hindering my overall work !

Need to find particular solution.
y"+6y'+5y=-9te^(5t)

What would be my guess work: I tried y_p=Ate^(5t)+Bte^(5t)
Any thought on the guess?
Thanks,
Dmitriy.

 

[dextercioby]

If you like lenghty calculations, you might use the method of variation of constants. Assume for example that the particular solution to the given ODE is of the form

y_{p}=C(t)e^{-5t}

You need to find the exact expression of C(t). For that plug in the ODE and then get another nonhomogenous second ODE whose particular solution you need to find

I get something like

C''(t)-4C'(t)=-9te^{10t}

for which i use the ansatz

C(t)=D(t)e^{4t}

and finally getting the ODE

D''(t)=-9te^{6t}

which can be integrated twice to get a solution depending on 2 constants of integration which you'll determine by plugging in the initial ODE.

Daniel.

 

[dextercioby]

It's too complicated what i did above (perhaps even incorrect).

Try the ansatz

y_{p}=C(t)e^{5t}

THe ODE for C(t) is

C''(t)+16C'(t)+60C(t)=-9t

whose particular solution is

C_{p}(t)=\frac{1}{25}-\frac{3}{20}t

and whose general solution, after using the ansatz leads to the same general solution of the initial ODE, so nothing new. Ergo, the particular solution to the initial ODE is

y_{p}(t)=\left(\frac{1}{25}-\frac{3}{20}t\right)e^{5t}

Daniel.

 

[Dimedrol]

dexterciby: This is works perfect.
however, it certainly interfere with my previous knowledge about non homogeneous EQ's. one of the solutions for the characteristic equation is e^(-5t), wouldn't we have to multiply everything by another "t"?
Please explain?

 

[dextercioby]

No, i absorbed any linear "t" dependence in the unknown function C(t).

Daniel.

 

[HallsofIvy]

Since e^5t is not a solution to the associated homogeneous equation (e^-5t is), a solution to the entire equation will be of the form y_p(t)= (At+ B)e^5t. Plug that into the equation and solve for A and B.