Solving some inhomogeneous differential equations I am currently reviewing some differential equations and ran into a couple of problems with the problems shown below particularly in the form of the particular solution for the equations. I haven't done this stuff in a couple years so I am pretty rusty.


*

*$u"-u'=6 +e^{2t}$


This one is giving me issues because the guess for the particular solution $(a + b e^{2t})$ loses the constant coefficient in the derivatives, and multiplying by $t$ (in case the homogeneous solution matches the particular form guess, which in this case it doesn't) , doesn't seem to work either.


*

*$u'+ u = 4e^{-t}$


This equation does have a matching particular/homogeneous solution, but when we multiply by $t$ for another unique solution it yields no positive results. 
Let me know what you think. Thanks
 A: Hint: I am assuming the DE is $u''-u=6+e^{2t}$. If that is the case, and you look for a particular solution of shape $at^2+be^{2t}$, you will soon find one. 
Note that in general if you have a linear inhomogeneous equation with the right-hand side equal to $f(t)+g(t)$, to get a particular solution all you need to do is to find a particular solution for right-hand side $f(t)$, one for $g(t)$, and add. 
A: The second equation can be straight up solved by integrating factors.
$$e^tu'+e^tu=(e^tu)'=4$$
$$e^tu=4t+k,u=4te^{-t}+ke^{-t}$$
The first can be solved similarly with the integrating factor $e^{-t}$ to obtain $(e^{-t}u')'$ on the left side.  From there, it should be straightforward.
A: Review: Writing $D$ to mean differentiation, if you have an equation
$$
     (D-c)f = 0
$$
then $f = Ae^{cx}$ for some constant $A$. If you have $(D-c)^{2}f=0$, then the possible solutions are
$$
        Ae^{cx}+Bxe^{cx}
$$
And these can be combined. For example, if the possible solutions of $(D-3)^{2}(D-7)^{3}$ are
$$
      (Ae^{3x}+Bxe^{3x})+(Ce^{7x}+Dxe^{7x}+Ex^{2}e^{7x})
$$
Your equation:
$$
    D(D-1)u = 6+e^{2t}
$$
The annihilator method tells you to apply a polynomial in $D$ that will kill the terms on the right. $D$ kills the $6$ and $(D-2)$ kills the $e^{2t}$ term. So you get a new equation
$$
                 D^{2}(D-1)(D-2)u = 0.
$$
The general solution is
$$
                   u(t)=(A+Bt)+Ce^{t}+Ee^{2t} \mbox{ (Can't use $D$ because it's used.)}
$$
(There's really an $e^{0t}$ multiplying $(A+Bt)$, but that's just $1$.)
The annihilator introduces extra freedom that isn't really there. So plug back into the original equation, and I think you'll find that $B=-6$ and $E=1/2$. The other constants can be arbitrary, which gives a final solution of
$$
        u(t) = A-6t+Ce^{t}+\frac{1}{2}e^{2t}.
$$
You expect a two-parameter family of solutions because the equation is second order.
