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Consider the non-homogeneous second order equation $t^2y''-3ty'+4y = t$.

Find a solution to the above problem of the form $y = t^r$ by direct substitution. How many solutions are there?

I have tried doing the substitution, finding $y'$ and $y''$, plugging it all in and I get $(r^2-4r+4)t^r = t$ but I am unsure what I am supposed to do from there.

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  • $\begingroup$ Solve the homogeneous problem then use variation of parameter to find a particular solution. $\endgroup$ – Jacky Chong Mar 14 '17 at 2:37
  • $\begingroup$ I have to use direct substitution, which is what is confusing me. The next question has us use variation of parameters. $\endgroup$ – user381985 Mar 14 '17 at 2:38
  • $\begingroup$ The substitution method is used to find homogeneous solutions. $\endgroup$ – Jacky Chong Mar 14 '17 at 2:39
  • $\begingroup$ Ok I get t^4*v'' + t^3*v' = t. This is pretty much the same place I got before. I'm not really sure what I am supposed to do from here. Sorry for me being stupid. $\endgroup$ – user381985 Mar 14 '17 at 2:49
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$$ t^2y'' -3ty' +4y = t $$ As Moo mentioned try $y = t^a v$ $$ y' = at^{a-1}v + t^a v'\\ y'' = a(a-1)t^{a-2}v + 2at^{a-1}v' + t^av'' $$ so we have $$ a(a-1)t^av + 2at^{a+1}v' +t^{a+2}v'' -3at^av - 3t^{a+1}v' +4t^a v = t $$ collecting terms $$ t^{a+2}v'' + \left[2a-3\right]t^{a+1}v' + \left[a(a-1) - 3a+4\right]t^a v = t^{a+2}v'' +\left[2a-3\right]t^{a+1}v' + \left[a^2-4a + 4\right]t^av = t $$ this shows that we could try to use $a = 2$ we find $$ t^4v'' +t^3v' = t\implies t^3v'' +t^2v' = 1 = 0 $$ we then use $u = v'$ we obtain a first order ode $$ t^3u' + t^2u = 1 $$ solve for $u$ then hope we can solve for $v$. and then we can sub back in for $y$.

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  • $\begingroup$ ps. Good luck. Side note just realized that you pretty much got to the solution in the comments. $\endgroup$ – Chinny84 Mar 14 '17 at 3:08

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