Differential equation with constant coefficients

Question

When we solve differential equation with constant coeffiecients we find out the auxiliary equation, then its roots and proceed further. But my question is When for example we have an equation of the type $ay'' +by=0$ auxiliary equation is $ar^2 +b=0$ but what if the differential equation is of the form $ay''+ by+c=0$ , then how do we write the auxiliary equation

Answer 1

A homogenous, second order differential equation with constant coefficients is of the form $ay'' + by' + cy = 0$, and are solved by solving the characteristic equation and plugging the solutions into $c_1 e^{r_1t} + c_2 e^{r_2t}$ (the formula is a bit different if you have a repeated root). Now, nonhomogenous equations are, in general, much more difficult to solve, even with constant coefficients. Here is a link to Paul's Notes that goes into the subject if you're interested:

http://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousDE.aspx

Here you'll find that there IS a link between the two, but you'll have to find a particular solution first. Since you have a constant on the right this should be fairly trivial.

Since I have some time on my hands, here is an example. Say we want to solve $y'' - 5y = 2$. We first see that the associated characteristic equation is $r^2 - 5 + 0 \implies r=\pm \sqrt{5}$ making our complementary solution $c_1 e^{t\sqrt{5}} + c_2 e^{-t\sqrt{5}}$
We now need a particular solution. Since we have a constant on the right side of our original equation our guess will be a constant, say $A$. We now plug this into the equation to get $A'' - 5A = 2 \implies 5A = -2 \implies A=-\frac{2}{5}$ (note that the nth derivative of a constant is a zero)
Now that we have found all this, we use the the fact that the solution set to a non-homogenous equation of this type is the sum of its complementary solution and a particular solution, which in this case is simply $c_1 e^{t\sqrt{5}} + c_2 e^{-t\sqrt{5}} -\frac{2}{5}$. At this point you can plug in any initial values, boundary conditions, etc.