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Find the general solution of:

$$ t^2y''+4ty'+y=0 $$

I am familiar with how to solve by constant coefficients for homogeneous second order DE's. However, I am not sure how to work with variable coefficients. Could someone point me in the correct direction? Thank you!

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    $\begingroup$ Hint: let $y = t^m$ - it is a Euler-Cauchy type of DE. $\endgroup$ – Moo Mar 13 '16 at 23:45
  • $\begingroup$ These are known as Euler Equations and their solutions are described in Paul's Online Math Notes $\endgroup$ – Mufasa Mar 13 '16 at 23:59
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Beside the fact that these are known differential equation (Euler-Cauchy family), what you could notice is that the equation is "homogeneous" in the sense that they are equidimensional, that is to say that they express as linear combinations of $t^n y^{(n)}$.

So, as Moo commented, the trick is to set $$y=t^m\implies y'=mt^{m-1}\implies y''=m(m-1)t^{m-2}$$ Replacing and dividing by $t^m$, you then arrive to an equation in $m$.

For the case you address, the resulting equation is then $$m(m-1)+4m+1=m^2+3m+1=0$$ simple to solve (a quadratic with roots $m_1$ and $m_2$).

This makes the solution to be $$y=c_1 t^{m_1}+c_2 t^{m_2}$$

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