Diff. Eq. — Identifying Particular Solution Given solution family

In summary: 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.
  • #1
WyattKEllis
2
0
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!
 
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  • #2
Are we assuming the the ODE has constant coefficients? If so [itex]t^n e^{kt}[/itex] will be a homogenous solution if and only if [itex]k[/itex] is a root of the characteristic equation with multiplicity [itex]n[/itex].

But here there is only one arbitrary constant, which suggests that only that one term is the homogenous solution.
 
  • #3
pasmith said:
Are we assuming the the ODE has constant coefficients? If so [itex]t^n e^{kt}[/itex] will be a homogenous solution if and only if [itex]k[/itex] is a root of the characteristic equation with multiplicity [itex]n[/itex].

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.
 
  • #4
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.
 
  • #5
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 [itex]y = y_h + y_p[/itex].
 

1. What is a particular solution in differential equations?

A particular solution in differential equations is a specific solution that satisfies both the differential equation and any initial conditions given. It is unique for a given set of initial conditions and can be found by substituting the initial conditions into the general solution.

2. How do you identify a particular solution given a solution family?

To identify a particular solution given a solution family, you need to use the initial conditions given in the problem. Substitute these initial conditions into the general solution and solve for the constants. The resulting equation will be the particular solution.

3. What is the difference between a general solution and a particular solution?

A general solution is a solution that contains an arbitrary constant or constants, while a particular solution is a specific solution that satisfies both the differential equation and any initial conditions given. The general solution represents all possible solutions to the differential equation, while the particular solution represents a single solution.

4. Can a particular solution be found without initial conditions?

No, a particular solution cannot be found without initial conditions. The initial conditions are necessary to determine the values of the constants in the general solution, which in turn gives the particular solution. Without initial conditions, the particular solution cannot be uniquely determined.

5. What happens if the initial conditions do not have a unique solution?

If the initial conditions do not have a unique solution, then the particular solution cannot be uniquely determined. This means that there are multiple solutions that satisfy the differential equation and initial conditions. In this case, the problem may need to be reevaluated or further information may be needed to determine the particular solution.

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