Question #95164

#(d^2y)/(dt^2)− y = 2e^t qquad star#

The general solution (#y_g#) is made up of the complementary (#y_c#) and particular (#y_p#) solutions.

The complementary solution is the solution to homogeneous DE:

#(d^2y)/(dt^2)− y = 0#

This has characteristic equation #lambda^2 - 1 = 0#, #lambda = pm 1#

And so #y_c = A e^t + B e^(-t)#

For the particular solution, we might think that #y_g = alpha e^(t)# is the way to go but that term is already in the complementary solution and so we look instead at #y_p = alpha t e^t#

So:

#y_p = alpha t e^t#

#y_p' = alpha e^t + alpha t e^t = alpha e^t(1 + t )#

#y_p'' = alpha e^t (1 + t ) + alpha e^t = alpha e^t (2+t)#

Plugging that into #star# yields:

#alpha e^t (2+t) - alpha t e^t = 2 e^(t) implies alpha = 1#

So the general solution is:

#y_g = A e^t + B e^(-t) + te^t#

First, solve for #(d^2y)/dt^2-y=0# to obtain the solution of the associated homogeneous equation. This gives #c_1e^t+c_2e^(-t)#.

Now try to solve #(d^2y)/dt^2-y=2e^t#. Guess that a particular solution #y# is of the form #Ae^t#, where #A# is a constant. Thus, #Ae^t-Ae^t=2e^t#, which is clearly not possible. Modify our guess to #Ate^t#. Thus, #A(te^t+e^t)+Ae^t-Ate^t=2e^t#, or #A=1#. This means that #te^t# is one of the solutions. Now, using the solutions of the associated homogenous equation, all solutions #y(t)# of this differential equation will satisfy #y(t)-te^t=c_1e^t+c_2e^(-t)#. Thus, all solutions satisfy #y(t)=te^t+c_1e^t+c_2e^(-t)#.