# Diff. Eq. — Identifying Particular Solution Given solution family

• WyattKEllis
So in summary, the given conversation discusses finding the solution to a first order linear differential equation with constant coefficients, using the characteristic equation and homogenous and particular solutions. The solution is given by y(t) = y_h(t) + y_p(t), where y_h(t) is the homogenous solution and y_p(t) is the particular solution. The forcing function is C*(e^t)+((t^3)+(2t^2)+2t-1)*(e^t)+((2t^3)+(t^2)+8t-7)*(e^2t) and the associated homogeneous differential equation is y'_h+p(t)*y_h=0.

#### WyattKEllis

Homework Statement
Identify the homogenous solution and the particular solution. C is an arbitrary constant.
Relevant Equations
y(t) = C*(e^t)+((t^3)-(t^2)+5t-6)*(e^t)+((t^3)-(t^2)+5t-6)*(e^2t)
I identified the root 1 with multiplicity 1 and the root 2 with multiplicity 1. So The characteristic equation is ((m-1)^2)*(m-2)=0. Simplifying and substituting with y I found: y'''-4y''+5y'-2y=0.

So now I've realized that this is actually describing y(t)=(C1)*(e^t)+(C2)*(e^t)+(C3)*(e^2t) and I do not know how to proceed accounting for the term ((t^3)-(t^2)+5t-6) included twice. In general I don't know of another step to begin with and haven't had luck finding any instruction on how to move from a solution family to homogenous or particular solutions.

I'd appreciate some hint as to where to start; thanks!

Are we assuming the the ODE has constant coefficients? If so $t^n e^{kt}$ will be a homogenous solution if and only if $k$ is a root of the characteristic equation with multiplicity $n$.

But here there is only one arbitrary constant, which suggests that only that one term is the homogenous solution.

pasmith said:
Are we assuming the the ODE has constant coefficients? If so $t^n e^{kt}$ will be a homogenous solution if and only if $k$ is a root of the characteristic equation with multiplicity $n$.

But here there is only one arbitrary constant, which suggests that only that one term is the homogenous solution.

Thank you! I was definitely on the wrong track. Now I've identified y_h(t)=C*(e^t) as the homogenous solution and y_p(t)=((t^3)-(t^2)+5t-6)*(e^t)+((t^3)-(t^2)+5t-6)*(e^2t) as the particular solution which I believe is appropriate.

So this solution family satisfies some first order linear differential equation which I would now like to find. I found this to be dy/dt=C*(e^t)+((t^3)+(2t^2)+2t-1)*(e^t)+((2t^3)+(t^2)+8t-7)*(e^2t) by simply differentiating the given solution family.

To be as descriptive as possible should anyone wish to correct me, I found the standard form coefficient function to be 0 and the forcing function to be C*(e^t)+((t^3)+(2t^2)+2t-1)*(e^t)+((2t^3)+(t^2)+8t-7)*(e^2t). Then I applied what the problem stated as the associated homogeneous differential equation y'_h+p(t)*y_h=0 and found that this equation is only true when C=0. This seems backwards (since this was mentioned as a way to check the answer) but is otherwise the best attempt I could come up with.

WyattKEllis said:
y'_h+p(t)*y_h=0

I assume you have to find p(t) such that Ce^t is a solution to the homogenous first order differential equation.

what does the notation _h mean? I have not seen it in other ODE texts.

docnet said:
I assume you have to find p(t) such that Ce^t is a solution to the homogenous first order differential equation.

what does the notation _h mean? I have not seen it in other ODE texts.
I assume the homogenous solution, with $y = y_h + y_p$.