Cauchy-Euler equation/Proofs


 * Main article: Cauchy-Euler equation

This article will cover all of the necessary technical proofs that could not be displayed in the main article. If you're looking for the computational method, please see the main article instead.

Repeated real root
Since $$y_1 = x^m$$ is a solution to the homogeneous differential equation, then we will apply reduction of order to find the second.

Assume that the second solution is of the form
 * $$y_2 = f(x) y_1$$, where $$f(x)$$ is a function of $$x$$.

Then differentiation gives
 * $$y_2' = f'(x) y_1 + f(x) y_1', \quad y_2 = f(x) y_1 + 2f'(x) y_1' + f(x) y_1''$$.

Substituting the derivatives into the differential equation gives
 * $$x^2 \left(f(x) y_1 + 2f'(x) y_1' + f(x) y_1\right) + x\left(f'(x) y_1 + f(x) y_1'\right) + f(x) y_1 = 0$$.

Factoring out $$f(x)$$, $$f'(x)$$ and $$f''(x)$$ yields
 * $$f(x)\left(x^2 y_1 + xy_1' + y_1\right) + f'(x)\left(2x^2 y_1' + xy_1\right) + f(x)(y_1) = 0$$.

Now, since $$y_1$$ is a solution to the homogeneous differential equation, it then follows that
 * $$x^2 y_1'' + xy_1' + y_1 = 0$$.

So the differential equation simplifies to
 * $$f'(x)\left(2x^2 y_1' + xy_1\right) + f''(x)(y_1) = 0$$.

Let $$v(x) = f'(x)$$, which then reduces the differential equation into a first order differential equation
 * $$v(x)\left(2x^2 y_1' + xy_1\right) + v'(x)(y_1) = 0$$.

This differential equation can be solved in a number of ways; we shall solve this using the integrating factor. We will rewrite the differential equation as
 * $$v'(x) + v(x)\left(\frac{2x^2 y_1' + xy_1}{y_1}\right) = 0$$.

We shall attempt to evaluate
 * $$\int \frac{2x^2 y_1' + xy_1}{y_1}\,dx$$.

By splitting up the fraction, we arrive at
 * $$\int \frac{2x^2 y_1' + xy_1}{y_1}\,dx = \int \left(\frac{2x^2 y_1'}{y_1} + x\right)\,dx$$.

By writing $$y_1$$ explicitly as $$y_1 = x^m$$, we arrive at
 * $$\int \frac{2x^2 y_1' + xy_1}{y_1}\,dx = \int \left(2mx + x\right)\,dx$$

Integrating, we get
 * $$\int \left(2mx + x\right)\,dx = mx^2 + \frac{1}{2}x^2.$$

So the integrating factor is
 * $$I = e^{mx^2 + \frac{1}{2}x^2}$$.

Multiplying both sides of the differential equation by the integrating factor and simplifying, we end up with the simple equation
 * $$\left(e^{mx^2 + \frac{1}{2}x^2}\right)\left(v'(x) + v(x)\left(\frac{2x^2 y_1' + xy_1}{y_1}\right)\right) = 0 \Rightarrow \frac{d}{dx}\left(v(x) e^{mx^2 + \frac{1}{2}x^2}\right) = 0.$$

Integrating both sides give us
 * $$v(x) \left(e^{mx^2 + \frac{1}{2}x^2}\right) = C \implies v(x) = Ce^{-\left(mx^2 + \frac{1}{2}x^2\right)}.$$

Finally, integrating both sides gives