Given 3 solutions to a second order non-homogeneous differential equation, find the general solution. We are given: \[\psi_1(t) = t^2, \psi_2(t)= t^2+e ^{2t}, \psi_3(t)= 1+t^2+2e ^{2t}\] Now my book gives the lemma that the difference of any two solutions to the non-homogeneous equation are in turn solutions to the homogeneous equation. So, my first step clearly is to take some differences.
I chose to do \[\psi_1-\psi_2 = e ^{2t}\] and \[\psi_3 - \psi_2 = 1+e ^{2t}\]
Now, from what I understand in my book, the general solution should be of the form \[y(t) = c_1y_1(t)+c_ty_2(t)+\psi(t)\] so, using those differences, I have \[y(t)=c_1(1+e ^{2t})+c_2e ^{2t}+ \psi_1\]
but the book says the solution to this is \[y(t) = c_1 + c_2e ^{2t}+ t^2\] Does anyone see what I am missing here? I dont see how any difference I take will result in this.. please help!
@hartnn @Hero would either of you mind taking a look at this? I don't think many people here can help with this one. I am pretty sure there is something I am missing here that is simple but not obvious to me.
c1 and c2 are arbitrary constants that you can solve for later, right?
Then what you have gives you have \[y(t) = c_1(1+e^{2t}) + c_2e^{2t}+t^2\] \[y(t) = c_1 + c_1e^{2t} + c_2e^{2t}+t^2\] \[ y(t) = c_1 + (c_1+c_2)e^{2t} + t^2\] then just call that quantity of constants your new c_2 \[y(t) = c_1 + c_2e^{2t} + t^2\] magics! ^_^
Join our real-time social learning platform and learn together with your friends!