# Euler Differential Equation General Solution

Given the initial value problem of Euler Differential Equation: $$x^2y''+\beta xy'+\alpha y=0$$ $$y(-1)=2 , y'(-1)=3$$

According to my book, the general solution for x<0 is the same as that of x > 0 so all the possible general solutions should be expressed with absolute value of x. For example, if we assume $y= x^r$ then if the characteristic equation resulted in a repeated root for r, then the general solution would be $y= c_1\left | x \right |^r+ c_2\left | x \right |^rln(\left | x \right |)$ with the absolute value for both x terms. However, when I encountered an example where the initial conditions were at a negative x point, my book expressed the general solution without the absolute value of x which gave a different answer from the one with the general solution expressed with the absolute value of x. So which is correct, expresseing the general solution with or without the absolute values?

## 1 Answer

To explain, we must go back to why the trial solution $y=x^r$ works here. Substitute $x = e^t$ to get

$$\frac{d^2y}{dt^2} + (\beta-1)\frac{dy}{dt} + \alpha y = 0$$

This is a linear equation with constant coefficients, therefore the general solution has the form $y(t) = e^{rt}$, which leads to $y(x) = x^r$. A double root gives $y = c_1e^{rt} + c_2 t e^{rt} = c_1 x^r + c_2 x^r \ln(x)$

In both cases, the absolute value is not needed, because our original substitution $x=e^t$ assumes $x > 0$. However, if an initial condition is given for $x < 0$, this assumption falls apart, and we must instead substitute $x = -e^t$. Fortunately, this substitution results in the same equation in $t$, and we get the general solution $y = e^{rt} = (-x)^r$, or $y = te^{rt} = (-x)^r\ln(-x)$

So to answer your question, which form of the solution to use depends on the context of the problem. If an initial condition is given, either the one with $(x)$ or $(-x)$ is sufficient. If no initial condition is given, then using the absolute value on $|x|$ would encompass both cases, giving the most general form. In practice, the absolute value is often ignored.

You might notice that a specific solution cannot include $x=0$ in its domain, due to it being a singular point. Hence, if an initial condition is given in $x<0$, then the solution cannot exist in $x>0$, and vice versa.