Solving differential equation (t^2-1)y''-6y=1

  • #1
CGandC
326
34
Homework Statement
Solve the ## \mathrm{DE}\left(t^2-1\right) \ddot{y}-6 y=1## if it is known that the corresponding homogeneous problem has a polynomial particular solution.
Relevant Equations
I haven't learned about frobenius method.
I've learned about variation of parameters and I think that's the key here.
This was a question from some past exam, I found online a solution but it uses Frobenius method which wasn't taught in the course.

I would approach the solution by attempting to find a solution to the homogeneuous DE ##\left(t^2-1\right) \ddot{y}-6 y=0##, but that by itself is quite tricky. Using the hint, I've found a homogeneuous solution ## y_{homogeneuous_1} = t- t^3 ##, and a particular solution ## y_{particular} = -1/6 ## to the non-homogeneuous equation.



However, I'm unable to find a general solution which is of the form ## y = c_1\cdot y_{homogeneuous_1} + c_2 \cdot y_{homogeneuous_2} + y_{particular} ## since I lack another homogeneuous solution ## y_{homogeneuous_2} ## ( which should be linearly independent to ## y_{homogeneuous_1} ## ) which I can't find; I tried approaching the solution to the last problem using the fact ( variation of parameters ) that ## y_{particular} = a_1(x)\cdot y_{homogeneuous_1} + a_2(x) \cdot y_{homogeneuous_2} ##, but I got stuck.



Thanks in advance for the help!
 
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  • #2
If [itex]y_1[/itex] is a solution of the homogenous equation [tex]
y'' + py' + q = 0[/tex] then [tex]
y_2 = y_1 \int \frac{W}{y_1^2}\,dx[/tex] is a linearly independent solution, where the Wronskian [itex]W[/itex] satisfies [itex]W' + pW = 0[/itex]. Since here [itex]p = 0[/itex] you can take [itex]W = 1[/itex].
 
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Likes CGandC
  • #3
Thank you sir! I understand now
 

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