Cauchy-Euler Equation

Consider 2nd-order linear differential equation with variable coefficients.

$$a_{2}x^{2}y^{\prime\prime} + a_{1}xy^{\prime} + a_{0}y = f(x) \ (x > 0)$$

Let $x = e^{t}$. Then for $x > 0$, we have $t = \log{x}$ and

$$\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = \frac{1}{x}\frac{dy}{dt},$$

Thus

$$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\frac{1}{x})\frac{dy}{dt} + \frac{1}{x}\frac{d}{dx}(... ... \frac{1}{x^{2}}\frac{dy}{dt} + \frac{1}{x}(\frac{d^{2}y}{dt^{2}}\frac{dt}{dx})$$

$$= - \frac{1}{x^{2}}\frac{dy}{dt} + \frac{1}{x}(\frac{d^{2}y}{dt^{2}}\frac{1}{x}) = \frac{1}{x^{2}}(\frac{d^{2}y}{dt^{2}} - \frac{dy}{dt}).$$

From this,

$$x\frac{dy}{dx} = \frac{dy}{dt}, \ x^{2}\frac{d^{2}y}{dx^{2}} = \frac{d^{2}y}{dt^{2}} - \frac{dy}{dt}$$

Put these back to the original equation. We have the following linear differential equation with the constant coefficients.

$$a_{2}(\frac{d^{2}y}{dt^{2}} - \frac{dy}{dt}) + a_{1}\frac{dy}{dt} + a_{0}y = f(e^{t}).$$

The characteristic equation of the Cauchy-Euler equation is given by

$$a_{2}m^{2} + (a_{1} - a_{2})m + a_{0} = 0$$

and is called the indicial equation.

There is a simple way to find the indicial equation. Substitute $y = x^m$ into the following differential equation.

$$a_{2}x^{2}y^{\prime\prime} + a_{1}xy^{\prime} + a_{0}y = 0 (x > 0)$$

Then we have

$$a_2x^2 m(m-1)x^{m-2} + a_1 xmx^{m-1} + a_0 x^m = (a_2 m^2 + (a_1-a_2)m + a_0)x^m = 0$$

Example 2..18   Solve the following differetial equation

$$x^{2}y^{\prime\prime} - 2xy^{\prime} + 2y = 0 \ (x>0)$$

SOLUTION Let $x = e^{t}$ and $y = x^{m}$. Then the indicial equation is $m(m-1) - 2m + 2 = 0$. Thus $m = 1,2$. Furtheremore, the indicial equation is the characteristic equation of

$$\frac{d^{2}y}{dt^{2}} - 3\frac{dy}{dt} + 2y = 0$$

The general solution is

$$y = c_{1}e^{t} + c_{2}e^{2t} = c_{1}x + c_{2}x^{2}\ensuremath{\ \blacksquare}$$

Example 2..19   Solve the following differential equation

$$x^{2}y^{\prime\prime} - 2xy^{\prime} + 2y = x^3 \ (x>0)$$

SOLUTION The example above, we found the complementary solution is $y_{c} = c_{1}e^{t} + c_{2}e^{2t}$. Now since $f(t) = e^{3t}$, we find the particular solution by the method of undetermined coefficients. Substitute $y_{p} = Ae^{3t}$ into

$$\frac{d^{2}y}{dt^{2}} - 3\frac{dy}{dt} + 2y = e^{3t}$$

Then

$$2Ae^{3t} = e^{3t}$$

Thus, $A = \frac{1}{2}$ and the general solution is

$$y = c_{1}e^{t} + c_{2}e^{2t} + \frac{1}{2}e^{3t}$$

Now since $x = e^{t}$, we have

$$y = c_{1}x + c_{2}x^2 + \frac{x^3}{2}\ensuremath{\ \blacksquare}$$