Solving Euler Cauchy Equation: Y1,Y2, X^2 & More

[Hussam Al-Tayeb]

If for example, I have.
Y1=Cos(2Ln(x)) and Y2=Sin(2Ln(x)) and I have to reach the general solution.

I know how to get to the general solution is the cauchy equation:
y'' X^n + y' x + 4 y = 0
According to the answer, n=2 => y'' X^2 + y' x + 4 y = 0
How am I to know that that it is X^2 ?
Or is it always X^2 in a Euler Cauchy equation?
Thanks in advance.

 

[HallsofIvy]

If for example, I have.
Y1=Cos(2Ln(x)) and Y2=Sin(2Ln(x)) and I have to reach the general solution.

I know how to get to the general solution is the cauchy equation:
y'' X^n + y' x + 4 y = 0
According to the answer, n=2 => y'' X^2 + y' x + 4 y = 0
How am I to know that that it is X^2 ?
Or is it always X^2 in a Euler Cauchy equation?
Thanks in advance.

An "Euler-Cauchy" equation (often called an "equipotential" equation) is the simplest kind of linear equation with variable coefficients. Part of the definition says that the variable derivative of each derivative is simply a power of x with the exponent being the same as the order of the derivative. So, yes. The function multiplying the second derivative MUST be some constant times x². (Surely the book does not give it as "X^2"- textbook authors do no usually interchange small letters and capital letters!) Since you are given two independent solutions, the differential equation they solve will be second order. The simplest way to solve that problem would be to write the very general x^2y" + Bxy' + Cy= 0, plug in the two given solutions and you will have two linear equations for B and C.
ax^2 y'+ b

I responded, two days ago, thinking the problem was to find the differential equation. Rereading it I see that you wanted to know how to find the "general solution".

That's much easier. The set of all solutions to any "homogenous linear n^th order differential equation" form a vector space of dimension n. In this case, the equation if of order 2 so the "solution space" has dimension 2. That means that any set of 2 "independent vectors" (independent solutions) forms a basis and so any vector (solution) can be written as a linear combination of the two solutions. If cos(2ln(x)) and sin(2ln(x)) are independent (they are, but you should show that, perhaps by looking at the Wronskian) then any solution can be written in the form C₁cos(2ln(x))+ C₂sin(2ln(x)) (the "general solution").

 

[tacman]

In order to solve the Euler Cauchy equation, we need to first rewrite it in the standard form:
y'' + P(x)y' + Q(x)y = 0
In your example, we have y''x^2 + y'x + 4y = 0. To get it in the standard form, we can divide both sides by x^2:
y'' + (1/x)y' + (4/x^2)y = 0
Now we can see that P(x) = 1/x and Q(x) = 4/x^2.

To solve this equation, we can use the method of reduction of order. Let's assume that one solution is y1 = cos(2ln(x)). We can then use the method of variation of parameters to find the second solution, y2. This will give us the general solution in the form of y = c1y1 + c2y2.

To answer your question, no, it is not always x^2 in a Euler Cauchy equation. It depends on the specific equation and its coefficients. In some cases, the power of x may be different, and in some cases, it may not even be a polynomial function of x. It is important to first rewrite the equation in the standard form before solving it.

I hope this helps clarify the process of solving Euler Cauchy equations. Keep practicing and you will become more comfortable with them. Good luck!