# Second Order Homogeneous Equation

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!

• Hint: let $y = t^m$ - it is a Euler-Cauchy type of DE. – Moo Mar 13 '16 at 23:45
• These are known as Euler Equations and their solutions are described in Paul's Online Math Notes – Mufasa Mar 13 '16 at 23:59

## 1 Answer

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 $$y=c_1 t^{m_1}+c_2 t^{m_2}$$