Cauchy-Euler Equation

The Cauchy-Euler Equation plays a significant role in the theory of linear differential equations because of its direct application to Fourier’s method in deconstructing PDE’s (partial differential equations). Let’s learn how to write the Cauchy-Euler equation and how to solve various types of Cauchy-Euler equations here in this article. Also, get the solved examples on the Cauchy-Euler equation here.

Cauchy-Euler Equation General Form

The differentiation equation gives the Cauchy-Euler differential equation of order n as

Cauchy Euler equation 1

Here, ai; i = 1, 2, 3,…, n are constants and an ≠ 0.

Learn: Differential equations

Second Order Cauchy-Euler Equation

The second-order Cauchy-Euler equation is of the form:

Cauchy Euler equation 2

(or)

Cauchy Euler equation 3

When g(x) = 0, then the above equation is called the homogeneous Cauchy-Euler equation.

In mathematics, the most commonly used Cauchy-Euler equation is the second-order Cauchy-Euler equation.

Cauchy-Euler Equation Examples

Below are a few examples of Cauchy-Euler equations of a different order.

  • x2 y′′ − 9xy′ + 25y = 0
  • y′′ + 9y = sec 3x
  • x2y′′ + 4y = 0
  • x3 y′′′ − 6x2 y′′ + 19xy′ − 27y = 0

Read more:

How to Solve the Cauchy-Euler Differential Equation?

Go through the steps given below to understand the method of solving the second order Cauchy-Euler differential equation.

Step 1: Let us assume that y = y(x) = xr be the solution of a given differentiation equation, where r is a constant to be determined.

Step 2: Fill the above formula for y in the differential equation and simplify. That means replace y with xr.

Step 3: Solve the obtained polynomial equation for r.

Step 4: For each obtained value of r, xr is a solution to the actual Euler equation.

If there is only one value for r, then y1(x) = xr is one solution to the differential equation, and we can reach the general solution by reducing order.

If there are two different real values for r, i.e., r1 and r2, then xr1, xr2 will be the fundamental set of solutions, whereas the general solution to the differential equation is y(x) = c1xr1 + c2xr2.

Cauchy-Euler Equation Solved Problems

Question 1:

Solve: x2 y′′ − 6xy′ – 18y = 0

Solution:

Given second order Cauchy-Euler equation is:

x2 y′′ − 6xy′ – 18y = 0

Let y = xr and substitute in the given differential equation.

x2 y′′ − 6xy′ – 18y = 0

x2[ xr ]′′ – 6x[ xr ]′ – 18( xr ) = 0

x2 [r(r − 1)xr−2] − 6x [r xr−1] – 18xr = 0

(r2 − r)xr − 6r xr – 18xr = 0

(r2 – r – 6r – 18)xr = 0

⇒ r2 – 7r – 18 = 0

⇒ r2 + 2r – 9r – 18 = 0

⇒ r(r + 2) – 9(r + 2) = 0

⇒ (r + 2)(r – 9) = 0

⇒ r = -2, r = 9

So, r = -2 and 9 are the two distinct possible values of r.

Therefore, the set of fundamental solution is {x-2, x9} and the general solution is y = c1x-2 + c2x9.

Question 2:

Solve the differential equation x2 y′′ − 7xy′ + 16y = 0 for x > 0.

Solution:

Given second order Cauchy-Euler equation is:

x2 y′′ − 7xy′ + 16y = 0

Let y = xr and substitute in the given differential equation.

x2 y′′ − 7xy′ + 16y = 0

x2[ xr ]′′ – 7x[ xr ]′ + 16( xr ) = 0

x2 [r(r − 1)xr−2] – 7x [r xr−1] + 16xr = 0

(r2 − r)xr − 7r xr + 16xr = 0

(r2 – r – 7r + 16)xr = 0

⇒ (r – 4)2 = 0

⇒ r = 4

Therefore, the solution of the given equation is y1(x) = x4.

Since we got only one solution, we cannot simply write out the general solution. Let’s perform the reduction of order method.

Now, let y(x) = x4 u(x)

Let us differentiate the above equation.

y′(x) = [x4u]′ = 4x3 u + x4 u′

Again differentiating, we get;

y′′(x) = [4x3 u + x4 u′] ′ = 12x2 u + 8x3 u′ + x4 u′′

Now, take the actual equation, i.e.,

x2 y′′ − 7xy′ + 16y = 0

Substitute y(x), y'(x) and y”(x) in this equation, we get;

x2[12x2 u + 8x3 u′ + x4 u′′] – 7x[4x3 u + x4 u′] + 16[x4 u] = 0

12x4 u + 8x5 u’ + x6u” – 28x4 u – 7x5 u’ + 16x4 u = 0

x6u” + [8x5 – 7x5]u’ + [12x4 – 28x4 + 16x4]u = 0

x6u” + x5 u’ = 0

Let u’ = v.

So, x6v’ + x5v = 0

x6(dv/dx) + x5v = 0

x6(dv/dx) = -x5v

(1/v) (dv/dx) = -x5/x6

(1/v) (dv/dx) = -1/x

Integrating on both sides, we get;

∫ (1/v) (dv/dx) dx = ∫(-1/x) dx

ln |v| = -ln |x| + c0

So, v = ± e− ln|x|+c0 = c2/x

Now,

u(x) = ∫u ′ (x) dx = ∫ v(x) dx = ∫ c2/x dx = c2 ln |x| + c1

Hence, the general solution of the differential equation is:

y(x) = x4 u(x) = x4 [c2 ln |x| + c1] = c1x4 + c2x4 ln |x|

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