1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.