Euler-Cauchy Equations

An Euler-Cauchy equation is

\begin{displaymath}(EC)\;\;\;\;\;\;\;\;x^2 y'' + b x y' + c y = 0\end{displaymath}

where b and c are constant numbers. Let us consider the change of variable

x = et.

Then we have

\begin{displaymath}\frac{dy}{dx} = e^{-t}\frac{dy}{dt},\;\;\;\mbox{and}\;\;\;
...{dx^2} = \left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right)e^{-2t}.\end{displaymath}

The equation (EC) reduces to the new equation

\begin{displaymath}\frac{d^2y}{dt^2} + (b-1)\frac{dy}{dt} + cy = 0\;.\end{displaymath}

We recognize a second order differential equation with constant coefficients. Therefore, we use the previous sections to solve it. We summarize below all the cases:
Write down the characteristic equation

\begin{displaymath}r^2 + (b-1)r + c =0 \;.\end{displaymath}

If the roots r1 and r2 are distinct real numbers, then the general solution of (EC) is given by

y(x) = c1 |x|r1 + c2 |x|r2.

If the roots r1 and r2 are equal (r1 = r2), then the general solution of (EC) is

\begin{displaymath}y(x) = \Big(c_1 + c_2 \ln\vert x\vert\Big)\vert x\vert^{r_1}.\end{displaymath}

If the roots r1 and r2 are complex numbers, then the general solution of (EC) is

\begin{displaymath}y = \Big(c_1 \cos(\beta \ln\vert x\vert)+ c_2 \sin(\beta \ln\vert x\vert)\Big)\vert x\vert^{\alpha}\end{displaymath}

where $\alpha = -(b-1)/2$ and $\beta = \sqrt{4c - (b-1)^2}/2$.

Example: Find the general solution to

\begin{displaymath}x^2y'' -xy'+ y = 0\;.\end{displaymath}

Solution: First we recognize that the equation is an Euler-Cauchy equation, with b=-1 and c=1.
Characteristic equation is r2 -2r + 1=0.
Since 1 is a double root, the general solution is

\begin{displaymath}y(x) = \Big(c_1 + c_2 \ln\vert x\vert\Big)\vert x\vert.\end{displaymath}

[Differential Equations] [First Order D.E.] [Second Order D.E.]
[Geometry] [Algebra] [Trigonometry ]
[Calculus] [Complex Variables] [Matrix Algebra]

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Author: Mohamed Amine Khamsi

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