# Find the general solution to x^2y'' + 5xy' + 4y = 0.

## Question:

Find the general solution to {eq}x^2y'' + 5xy' + 4y = 0. {/eq}

## Euler Differential Equation.

This differential equation has non constant coefficients which is in general quite challenging. However this is a special type called an Euler differential equation. If you try the solution

{eq}\displaystyle y = x^r {/eq}

Then you can convert the equation into a quadratic equation in r.

{eq}\displaystyle x^2 (x^r)'' + 5x (x^r)' + 4x^r = 0 {/eq}

{eq}\begin{align*} (r(r-1) + 5r + 4)x^r &= 0\\ r^2 + 4r +4 &= 0\\ (r+ 2)(r+2) &= 0 \end{align*} {/eq}

Since the root repeats for this type of equation, you have to multiply the second solution by a natural logarithm.

{eq}\displaystyle y = c_1 x^{-2} + c_2 (\ln x) x^{-2} {/eq}

The is analogous to multiplying by t when you have a repeated root in constant coefficient differential equations.

To see that you need the logarithmic term you can substitute

{eq}\displaystyle y = u(x)x^{-2} {/eq}

back into the equation and solve for u(x) = (\ln (x) +C)