# Solving a differential equation using Cauchy-Euler Method

In summary, the student is having difficulty solving a problem using the Cauchy-Euler method that they are supposed to solve. They are also having trouble understanding the derivative notation in their book.

## Homework Statement

Hey all, this is my first time posting here. I've used your help before, but have not actually ever had to post a question. Thanks for any help you can give, I am excited to finally join the community.

I am doing calculus homework and I am having trouble solving this problem using the Cauchy-Euler method that we're supposed to solve it with.

Solve: xy4 + 6ym = 0

It is the part about the 6ym that I am not sure what to do with.

## Homework Equations

andny / dxn ...a0y = 0

## The Attempt at a Solution

My book toldme to try and substitute y=xm into the derivative parts, and all of the other problems I've done work with that method. This is what I've done so far.

y=xm
y1=mxm-1
y2=m(m-1)xm-2
y3=m(m-1)(m-2)xm-3
y4=m(m-1)(m-2)(m-3)xm-4

I know how to treat the 4th derivative term; again, I just don't know what or how to think about the y^m term. I know the solution to the problem, as it is in the back of my book, I just don't know how to get there. Can anyone give me a prod in the right direction?
Thanks

I don't see any derivative in the original equation!

I'm sorry, our book uses notation so that y^4 is supposed to be the fourth derivative of y.
So, for example, y1=y´ and y^2 = y´´

I'm sorry, our book uses notation so that y^4 is supposed to be the fourth derivative of y.
So, for example, y1=y´ and y^2 = y´´

The usual notation is y^(2) (so we can tell the difference between that and the square of y). Anyway, you have found the first few y^(m) for y = x^n. If you don't see the pattern, I suggest you look instead at y^(m)/m! and see if you recognize the coefficients there.

RGV

I see that there is a relationship in that
y^(m)/m = (my^(m-1) + m(m-1)y^(m-2)...)/(m(m-1)(m-2)...) but I don't actually understand how to apply that relationship at all yet

Nevermind, I'm really sorry about this. I mistook the book for having written y^m when they really wrote y^´´´. I guess my new problem is poor eyesight. Thanks a lot for the help, I struggled trying to solve this for a solid 40 minutes.

## 1. What is the Cauchy-Euler method for solving differential equations?

The Cauchy-Euler method is a technique used to solve linear differential equations of the form ax^2y'' + bxy' + cy = 0, where a, b, and c are constants. It involves finding a general solution by assuming a solution of the form y = x^r and then solving for the values of r that satisfy the equation.

## 2. When is the Cauchy-Euler method most useful?

This method is most useful when dealing with differential equations that involve a polynomial function multiplied by a derivative or a function raised to a power.

## 3. What are the steps for solving a differential equation using the Cauchy-Euler method?

The steps for solving a differential equation using the Cauchy-Euler method are as follows:1. Assume a solution of the form y = x^r.2. Substitute y and its derivatives into the differential equation.3. Solve for the values of r that satisfy the equation.4. Use the values of r to form the general solution y = c1x^r1 + c2x^r2 + ... + cnx^rn.5. Use initial conditions, if given, to determine the specific values of the constants c1, c2, ..., cn.

## 4. Can the Cauchy-Euler method be used to solve non-homogeneous differential equations?

Yes, the Cauchy-Euler method can be used to solve non-homogeneous differential equations by first finding the general solution to the associated homogeneous equation and then using the method of undetermined coefficients to find a particular solution to the non-homogeneous equation.

## 5. What are the limitations of the Cauchy-Euler method?

The Cauchy-Euler method can only be used to solve linear differential equations and is not applicable to non-linear equations or systems of equations. Additionally, it may not always produce a real-valued solution, as it relies on the assumption of a solution in the form of a power function.

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