Euler-Cauchy Equations

An Euler-Cauchy equation is

$$(EC)\;\;\;\;\;\;\;\;x^2 y'' + b x y' + c y = 0$$

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

x = e^t.

Then we have

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

The equation (EC) reduces to the new equation

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

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:

(1)

Write down the characteristic equation

$$r^2 + (b-1)r + c =0 \;.$$

(2)

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.

(2)

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

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

(3)

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

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

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

Example: Find the general solution to

$$x^2y'' -xy'+ y = 0\;.$$

Solution: First we recognize that the equation is an Euler-Cauchy equation, with b=-1 and c=1.

1

Characteristic equation is r² -2r + 1=0.

2

Since 1 is a double root, the general solution is

$$y(x) = \Big(c_1 + c_2 \ln\vert x\vert\Big)\vert x\vert.$$

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

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