# This problem is hindering my overall work

• Dimedrol
In summary, Daniel is trying to solve for the constants A and B in order to find the particular solution to the given ODE. However, because the equation is nonhomogeneous, he has to use an alternative method.
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.

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.

Last edited:
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.

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"?

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

Daniel.

Since e5t is not a solution to the associated homogeneous equation (e-5t is), a solution to the entire equation will be of the form yp(t)= (At+ B)e5t. Plug that into the equation and solve for A and B.

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