Solving differential equation $x^2y''-xy'+y=0, x>0$ with non-constant coefficients using characteristic equation? Whenever you deal with non-constant coefficients you usually use Laplace transform to solve a given differential equation, at least that's how how I learned it.
But how would you solve the equation using the characteristic equation?
I've got $x^2r^2-xr+1=0$, but not sure where to go from here? I am asked to obtain one solution from the equation.
 A: Problem Version 1:
We are given:
$$\tag 1 x^2y''+y=0, ~~x \gt 0$$
This is a Euler-Cauchy type DEQ.
We can let $y = x^m$, so we have: $y'(x) = mx^{m-1}, ~~y''(x) = m(m-1)x^{m-2}$.
Substituting this back into $(1)$, yields:
$x^2y''+ y = x^2(m(m-1)x^{m-2}) + x^m = x^m(m^2 - m + 1) = 0$.
So, we have a characteristic equation:
$$m^2 - m + 1 \rightarrow m_{1,2} = \dfrac{1}{2} \pm \dfrac{i \sqrt{3}}{2}$$
Now, because we are given $x \gt 0$ (if not, all the $x$ terms would have absolute values), we can write:
$$\displaystyle y(x) = y_1(x) + y_2(x) = c_1 x^{m_1} + c_2 x^{m_2} = c_1 x^{\large \frac{1}{2} + \frac{i \sqrt{3}}{2}} + c_2 x^{\large \frac{1}{2} - \frac{i \sqrt{3}}{2}}$$
Using Euler's identity, some algebra, and the fact that $x \gt 0$, we can write this as:
$$y(x) = c_1 \sqrt{x} \cos \left(\frac{\sqrt{3}}{2} \ln x\right) + c_2 \sqrt{x} \sin\left(\frac{\sqrt{3}}{2} \ln x\right)$$
Problem Version 2:
$$\tag 2 x^2y''-x y'+ y = 0, x \gt 0$$
This requires the use of the Frobenius method to find a solution near $x = 0$. I will assume you are studying this, so I am going to map it out for you and you fill in the missing details.
Here, we want the form $y'' + P(x) y' + Q(x) y = 0$, so:

*

*Dividing both sides by $x^2$, we get $P(x) = -\dfrac{1}{x}, ~ Q(x) = \dfrac{1}{x^2}$

*So, $x = 0$ is a singular point and the method of Frobenius is usable.

Next, to use this method, we substitute $y$ and its derivatives from:

*

*$\displaystyle y = \sum_{n=0}^\infty a_nx^{\lambda+n}$

*$y' = \lambda a_0 x^{\lambda - 1} + \ldots (\lambda+n+1)a_{n+1}x^{\lambda+n}+\ldots$

*$y'' = \lambda(\lambda-1)a_0x^{\lambda-2} + (\lambda + n + 1)(\lambda+n)a_{n+1}x^{\lambda + n -1}+\ldots$
We substitute these into $(2)$, collect like terms and set them equal to zero. This gives us:
$$x^{\lambda}(\lambda-1)^2a_0 + x^{\lambda + 1} \lambda^2 a_1 + \ldots x^{\lambda + n}\left((\lambda + n)^2 - 2(\lambda + n) + 1\right)a_n + \ldots = 0$$
From this, we get:

*

*$\tag 3 (\lambda - 1)^2 a_0 = 0$ and generally

*$\tag 4 \left((\lambda + n)^2 - 2(\lambda + n) +1\right)a_n = 0$
The first expression is called the indicial equation and it gives us: $(\lambda-1)^2 = 0$, which has a double root $\lambda_{1,2} = 1$, which we substitute into the second equation, yielding $n^2 a_n = 0$, which implies $a_n = 0, ~ n \ge 1$, thus:
$$y_1(x) = a_0 x$$
Now, we want a second linearly independent solution, but the indicial equations roots are equal. This complicates matters! To find a second family of solutions, we first solve for $a_n$ in terms of $\lambda$, where $\lambda$ is thought of as an independent variable in $(4)$. We obtain $a_n = 0$ for $n \ge 1$ and then substitute back into the original $y$ we used, with $\lambda$ as the independent variable, so:
$$y = y(\lambda, x) = a_0x^\lambda$$
Another family of solutions to $(2)$ can be obtained by:
$$y_2 = \dfrac{\partial y(\lambda, x)}{\partial \lambda} ~\text{evaluated at}~ \lambda = 1$$
Since (recall that we are given $x \gt 0$, else we would need absolute values):

*

*$y = a_0x^\lambda = a_0 e^{\lambda \ln x}$

*$\dfrac{\partial y(\lambda, x)}{\partial \lambda} = a_0 \ln x e^{\lambda \ln x} = a_0 x^\lambda \ln x$ , and evaluated at $\lambda = 1$, yields: $a_0 x \ln x = y_1(x) \ln x$

*This gives us the second solution as $y_2(x) = a_0 x^\lambda \ln x$.

Putting this together, our solution is (test this by replacing it in the original DEQ and see that is satisfies it):
$$y(x) = y_1(x) + y_2(x) = c_1 x + c_2 x \ln x$$
Lastly, it is worth noting the Wolfram Alpha finds a solution using modified Bessel and the Gamma functions.
A: Transfer your equation with $x = e^t$
Then $\frac{dx}{dt} = e^t$
$$\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = e^{-t}\frac{dy}{dt} = x^{-1} \frac{dy}{dt}$$
So $$x \frac{dy}{dx} = \frac{dy}{dt}$$
Do the same for $x^2 \frac{d^2y}{dx^2}$ and you shall get the result
$$x^2 \frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt}$$
I have omited the calculation part and hope you can do it for yourself. Now put all these values to your differential equation and you shall get the following
$$\frac{d^2y}{dt^2} - \frac{dy}{dt} + y =0 $$
Now you know how to solve.
Please check the calculation part only.
