Overview:
In a regular second order linear homogeneous equation the solution is easy to solve for after learning it in class. The equation should usually be in the following form when solving:(homogeneous=0). Where A, B, and C are constants and the highest degree should be the 2nd.
Ay”+By+Cy=0
The next step would be to get the characteristic equation of the differential equation from above or manage to put it into this form:
r²+Br+C
From here we need to be able to solve for the characteristic equation. This can be done in two ways. Either by factoring the equation or if it is not factor-able you may use the quadratic equation.
After getting the roots of the equation; we will call them r1 and r2, you can now proceed to solving the equation. The general solution of the characteristic equation is:
y=Ae^(r1*t)+Be^(r2*t)
This case is true for normal characteristic equations where the roots are real and different from each other , but in the case of this topic we would like to look at repeated roots; this would be when r1=r2.
If we try to solve the homogeneous equation using normal means with the repeated root the answer would come up to zero; that is not the correct answer because the values we are trying to solve for get cancelled out.
So in order to solve a equation with repeated roots we need to do one thing which is add a “t” in front of one of the original terms in the solution. Meaning the new equation we need to plug in values to would be:
y=Ae^(r1*t)+tBe^(r2*t)
This would the final equation we use in order to solve an equation with repeated roots where A and B are the constants we want to solve for.
Example/Sample Problem:
Now lets try solving for a sample problem with repeated roots.
y”+18y’+81y=0 ; y(0)=-5, y'(0)=48
So the first step is to set up the characteristic equation:
r²+18r+81
Now we need to factor or solve for the values of r1 and r2.
(r+9)*(r+9)
r1=-9, r2=-9
Now as we can see the roots are exactly the same or repeated roots. So we have to plug it into the equation which would look like this:
y=Ae^(-9t)+tBe^(-9t)
The next step would be to solve for the derivative of y.
y’=-9Ae^(-9t)-9tBe^(-9t)+Be^(-9t)
Now that we have both y and y’ we can now plug in the inital values of y(0)=-5 and y'(0)=48 to solve for the values of A and B.
For y(o)>>> -5=Ae^(-9*0)+0*Be^(-9*0)
-5=A
For y'(0)>>> 48=-9Ae^(-9*0)-9*0Be^(-9*0)+Be^(-9*0)
48=-9A+B
48=-9(-5)+B
B=3
So now that we solved for both A and B we can now put it back into the original equation.
y=-5e^(-9t)+3te^(-9t)
This is the final answer for the repeated roots problem.
3. In case I wasn’t clear enough with the explanation or steps needed to solve the equation, or even if you just didn’t understand because I skipped some steps while solving the equation you can refer to these two YouTube links for more help. The videos could help to provide an extra problem that you can work out while watching the video. The YouTuber provides excellent visuals and a great explanation of how the problems should be solved.
(Disclaimer: Not owned by me, all credits go to “Khan Academy”)
