# Solving non-homogeneous second order differential equation

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.

• Solve the homogeneous problem then use variation of parameter to find a particular solution. – Jacky Chong Mar 14 '17 at 2:37
• I have to use direct substitution, which is what is confusing me. The next question has us use variation of parameters. – user381985 Mar 14 '17 at 2:38
• The substitution method is used to find homogeneous solutions. – Jacky Chong Mar 14 '17 at 2:39
• 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. – user381985 Mar 14 '17 at 2:49

$$t^2y'' -3ty' +4y = t$$ 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 $y$.