Linear differential equation - CauchyEuler Equations

Cauchy–Euler Equations Introduction Cauchy–Euler equations are a special class of differential equations that look different from the constant-coefficient linear equations you've studied before. The key distinguishing feature is that powers of the independent variable $x$ multiply the derivatives. Despite this apparent complexity, there's an elegant trick to solve them: a simple substitution transforms them into the familiar constant-coefficient form you already know how to solve. What is a Cauchy–Euler Equation? A Cauchy–Euler equation (also called an equidimensional equation) has the form: $$a0 x^n y^{(n)} + a1 x^{n-1} y^{(n-1)} + \cdots + a{n-1} x y' + an y = 0$$ where $a0, a1, \ldots, an$ are constant coefficients and $y^{(k)}$ denotes the $k$-th derivative of $y$ with respect to $x$. Key observation: Notice the pattern—the power of $x$ multiplying each derivative decreases by one as you move to lower derivatives. For example, the $n$-th derivative is multiplied by $x^n$, the $(n-1)$-th derivative by $x^{n-1}$, and so on. A common example is the second-order case: $$a x^2 y'' + b x y' + c y = 0$$ where $a, b, c$ are constants. The Solution Method: Substitution The remarkable fact about Cauchy–Euler equations is that they can be transformed into linear equations with constant coefficients using a substitution. The Key Substitution To solve a Cauchy–Euler equation, use the substitution: $$x = e^t \quad \text{or equivalently} \quad t = \ln x$$ This transforms the independent variable from $x$ to $t$, and it transforms the Cauchy–Euler equation into a constant-coefficient linear equation in $t$. Once you have that equation, you can solve it using the exponential trial method you already know. Why This Works When you change variables from $x$ to $t$ using $x = e^t$, something magical happens to the derivative terms. Using the chain rule: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1}{x} \frac{dy}{dt}$$ For the second derivative, applying the quotient and chain rules gives: $$\frac{d^2y}{dx^2} = \frac{1}{x^2}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right)$$ Notice that the denominators contain powers of $x$—exactly the powers we need to cancel the $x$ terms in the original Cauchy–Euler equation! When you substitute these expressions back into the original equation, all the $x$ terms cancel out, leaving only constant coefficients multiplying the derivatives of $y$ with respect to $t$. Summary of the Process Start with a Cauchy–Euler equation in terms of $x$ and $y$ Apply the substitution $x = e^t$ and use the chain rule to express $dy/dx$, $d^2y/dx^2$, etc. in terms of derivatives with respect to $t$ Substitute these expressions into the original equation to get a constant-coefficient equation in $t$ Solve the transformed equation using the exponential trial method Back-substitute $t = \ln x$ to express your solution in terms of $x$ <extrainfo> Example Application To see this concretely, consider solving $x^2 y'' + xy' - 4y = 0$. With the substitution $x = e^t$, the derivatives become: $\frac{dy}{dx} = \frac{1}{x}\frac{dy}{dt}$ $\frac{d^2y}{dx^2} = \frac{1}{x^2}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right)$ Substituting into the original equation: $$x^2 \cdot \frac{1}{x^2}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right) + x \cdot \frac{1}{x}\frac{dy}{dt} - 4y = 0$$ $$\frac{d^2y}{dt^2} - \frac{dy}{dt} + \frac{dy}{dt} - 4y = 0$$ $$\frac{d^2y}{dt^2} - 4y = 0$$ This is a standard constant-coefficient equation! You can now solve it by trying $y = e^{rt}$, which gives the characteristic equation $r^2 - 4 = 0$, so $r = \pm 2$. The general solution in terms of $t$ is $y = c1 e^{2t} + c2 e^{-2t}$. Converting back to $x$ using $t = \ln x$: $e^t = x$, so $e^{2t} = x^2$ and $e^{-2t} = x^{-2} = 1/x^2$. Therefore, the solution is $y = c1 x^2 + c2 x^{-2}$. </extrainfo>