Euler Form and Substitution Method
The Euler Form is a special variant of second-order linear ODEs, where there are extra factors of \( x \) and \( x^2 \) in front of the \( dy/dx \) and \( d^2y/dx^2 \) terms:
\begin{align}
x^2\frac{d^2y}{dx^2} + \alpha x \frac{dy}{dx} + \beta y = G(x) \tag{1}
\end{align}
This type of equation can be transformed into the usual second-order constant-coefficient ODEs by applying the substitution \( z = \ln x \). The derivatives now take the form of
\begin{align}
\frac{dy}{dx} &= \frac{dy}{dz}\frac{dz}{dx} \\
&= \frac{1}{x}\frac{dy}{dz} \tag{2} \\
\frac{d^2y}{dx^2} &= \frac{d}{dx}\left( \frac{dy}{dx} \right) \\
&= \frac{d}{dx} \left( \frac{1}{x}\frac{dy}{dz} \right) \\
&= \frac{d(1/x)}{dx} \frac{dy}{dz} + \frac{1}{x}\frac{d}{dx} \left( \frac{dy}{dz} \right) \\
&= -\frac{1}{x^2} \frac{dy}{dz} + \frac{1}{x} \frac{dz}{dx} \frac{d}{dz} \left( \frac{dy}{dz} \right) \\
&= -\frac{1}{x^2} \frac{dy}{dz} + \frac{1}{x} (\frac{1}{x}) \frac{d^2y}{dz^2} \\
&= -\frac{1}{x^2} \frac{dy}{dz} + \frac{1}{x^2} \frac{d^2y}{dz^2} \tag{3}
\end{align}
seen by using the Chain Rule. Substituting (2) and (3) into (1) then yields
\begin{align}
x^2\frac{d^2y}{dx^2} + \alpha x \frac{dy}{dx} + \beta y &= \tilde{G}(z) \\
x^2\left(-\frac{1}{x^2} \frac{dy}{dz} + \frac{1}{x^2} \frac{d^2y}{dz^2} \right) + \alpha x \left( \frac{1}{x}\frac{dy}{dz} \right) + \beta y &= \tilde{G}(z) \\
\frac{d^2y}{dz^2} + (\alpha -1) \frac{dy}{dz} + \beta y &= \tilde{G}(z) \tag{4} \\
\end{align}
where \( \tilde{G}(z) \) is the new source term after the transformation and we have converted the equation into the constant-coefficient form, which can be solved by the previous solution procedure.
Example
Find the general solution to the Euler form second-order ODE:
\begin{align}
x^2\frac{d^2y}{dx^2} + 4x\frac{dy}{dx} + 2y = \frac{1}{x} \tag{5}
\end{align}
By (4), the substitution \( z = \ln x \) (\( x = e^z \)) will transform the ODE into a constant-coefficient ODE:
\begin{align}
\frac{d^2y}{dz^2} + (4-1)\frac{dy}{dz} + 2y &= \frac{1}{e^z} \\
\frac{d^2y}{dz^2} + 3\frac{dy}{dz} + 2y &= e^{-z} \tag{6} \\
\end{align}
So we can proceed as before but with \( z \) as the new independent variable. The auxiliary equation is \( r^2 + 3r + 2 = 0 \) with roots of \( r = -1, -2 \), and thus \( y_c(z) = c_1 e^{-z} + c_2 e^{-2z} \). As suggested by this tutorial, to deal with the inhomogeneous part, we need to raise the degree due to duplication, and hence the ansatz for the particular solution will be \( y_p = Aze^{-z} \). Calculating the corresponding derivatives and plugging in, we have
\begin{align}
(-2Ae^{-z} + Aze^{-z})+ 3(Ae^{-z} -Aze^{-z}) + 2Aze^{-z} &= e^{-z} \\
(-2A+3A)e^{-z} &= e^{-z} \Rightarrow A = 1 \tag{7}
\end{align}
And hence the full general solution is
\begin{align}
y = y_c + y_p &= c_1 e^{-z} + c_2 e^{-2z} + ze^{-z} \\
&= \frac{c_1}{x} + \frac{c_2}{x^2} + \frac{\ln x}{x} \tag{8}
\end{align}
where we need to express it with the original \( x \).
Exercise
Solve the Euler form ODE in full:
\begin{align}
x^2\frac{d^2y}{dx^2} + 3x\frac{dy}{dx} + 2y = 0 \tag{9}
\end{align}
where the initial condition is \( y(1) = 1, y'(1) = 0 \).
Answer
Again, by the substitution \( z = \ln x \) like in (4), (9) will be turned into
\begin{align}
\frac{d^2y}{dz^2} + (3-1)\frac{dy}{dz} + 2y &= 0 \\
\frac{d^2y}{dz^2} + 2\frac{dy}{dz} + 2y &= 0
\end{align}
The auxiliary equation is \( r^2 + 2r + 2 = 0 \) with roots \( r_{\pm} = -1 \pm i \), and so the solution will be in the form of \( y(z) = c_1 e^{-z}\cos(z) + c_2 e^{-z}\sin(z) \). After a reverse transformation it looks like
\begin{align}
y(x) = c_1 \frac{\cos(\ln x)}{x} + c_2 \frac{\sin(\ln x)}{x}
\end{align}
Substituting the I.C., it is clear that \( c_1 = 1 \). And the first derivative is
\begin{align}
y'(x) = c_1 \frac{-\sin(\ln x)-\cos(\ln x)}{x^2} + c_2 \frac{\cos(\ln x)-\sin(\ln x)}{x^2}
\end{align}
so further plugging in the I.C. shows that \( c_2 = 1 \), \( y = (\cos(\ln x) + \sin(\ln x))/x \).
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