Non-homogeneous Linear Equations


Part d: Exponential Functions with a Conflict


Each differential equation has an exponential function on the right hand side.

The catch is that this exponential function is also a homogeneous solution. So the previous strategy of putting the function through and then dividing (or undetermined coefficients) won't work because you would need to divide by 0 (which you should never do!).

Each equation is factored for ease of calculation. You can use this to greatly simplify your calculations. The key is matching your power of e with the matching factor to reduce the power of t. Here are some examples of how that works.

(D−I)(tet)=et(D-I)(te^t) = e^t
(D+I)(te−t)=e−t(D+I)(te^{-t}) = e^{-t}
(D−3I)(te3t)=e3t(D-3I)(te^{3t}) = e^{3t}
(D−I)(t2et)=2tet(D-I)(t^2e^t) = 2te^t

Now here is how you can use it.

(D+3I)(D−2I)(te2t)(D+3I)(D-2I)(te^{2t})
=(D+3I)(e2t)= (D+3I)(e^{2t})
=2et+3e2t=5e2t.= 2e^t + 3e^{2t} = 5e^{2t}.

By putting the (D−2I)(D-2I) in front, te2tte^{2t} becamee2te^{2t}. Now it is a lot easier to do the rest.

Here is a higher order example.

(D+2I)(D−I)(D−I)(D−I)(t3et)(D+2I)(D-I)(D-I)(D-I)(t^3e^t)
=(D+2I)(D−I)(D−I)(3t2et)= (D+2I)(D-I)(D-I)(3t^2e^t)
=(D+2I)(D−I)(6tet)= (D+2I)(D-I)(6te^t)
=(D+2I)(6et)= (D+2I)(6e^t)
=(6et)+2(6et)= (6e^t)+2(6e^t)
=3(6et).= 3(6e^t).

Now it's your turn to use this to your advantage.



Find a particular solution for each equation.

No initial conditions are given because you are only asked to find one solution.


Please enter any multiplication explicitly with * symbols so the parcer can understand your answer, like this:

2t3e5t2⟶2∗t3∗e(5∗t2).2t^3e^{5t^2} \longrightarrow 2*t^3*e^{(5*t^2)}.

Problem 1

Find a particular solution for the differential equation below. The factored form is (D-3I)(D-2I).

y′′−5y′+6y=−e2ty'' - 5y' + 6y = -e^{2t}

Initial Conditions to Check: