Finding Solutions to 4th Order Differential Equations

  • Thread starter Thread starter Mathgician
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on solving the fourth-order differential equation xy^(4) + 6y'" = 0. The initial approach using the auxiliary equation m^4 + 6m^3 = 0 was incorrect due to the variable coefficient nature of the equation. The correct method involves recognizing the equation as an "Euler-type" equation by multiplying through by x^3, leading to a simpler separable first-order differential equation. The solution process includes substituting y''' with u and integrating three times to find the general solution.

PREREQUISITES
  • Understanding of fourth-order differential equations
  • Familiarity with auxiliary equations and their applications
  • Knowledge of Euler-type differential equations
  • Basic skills in solving separable first-order differential equations
NEXT STEPS
  • Study the method for solving Euler-type differential equations
  • Learn about variable coefficient differential equations
  • Explore the process of integrating higher-order differential equations
  • Review techniques for finding particular solutions to differential equations
USEFUL FOR

Students and educators in mathematics, particularly those studying differential equations, as well as anyone seeking to deepen their understanding of advanced calculus concepts.

Mathgician
Messages
78
Reaction score
0

Homework Statement


solve the given differential equation:

xy^(4) + 6y'" = 0

Homework Equations



No idea, I need something to get me started

The Attempt at a Solution



auxiliary equation:

m^4 + 6m^3 = 0

m^3(m+6)=0
3 roots with m = 0 and the other root is m = -6

so I got y = C1 + C2lnx + C3(lnx)^2 + C4x^-6

but, it is not the solution. Your help is greatly appreciated. Thank You.
 
Physics news on Phys.org
I don't quite know what you're doing here. The method you are using here seems to be the one that you use for a second order equation, however this is an equation of 4th order.

Take a look at the equation and try and spot a solution. In particular, you have a fourth derivative multiplied by x, and a third order derivative. Now, you know a function whose fourth derivative multiplied by x is equal to its third derivative (modulo constants). The function is xn. So, try the general solution y=xn. Plugging this into the ODE will give you a condition on n, which will enable you to obtain the particular solution.
 
Your auxiliary equation is wrong. You appear to have used the auxiliary equation for a "constant coefficients" equation when you have a variable coefficient- and then used the "Euler type" solution.

One way to handle this is to multiply then entire equation by x3 so that you get x4y(iv)+ 6x3y'''= 0, an "Euler-type" or "equipotential" equation. If you let y= xr you will get the correct auxilliary equation for this d.e.

Much simpler is to let u= y'''. Now the equation is xu'+ 6u= 0, a simple, separable first order differential equation. After you have solved that, let y'''= u= your solution and integrate three more times.
 
so the auxiliary method only works for 2nd order? Can you guys give me little more insight or any websites that would explain it to me further because my books was very vague in explain my problem or there were no example of such problem in the book. I hate my textbook. Thank you for your insights guys :shy:
 
can you give me hint on how to solve an equation that involves a particular solution? Thank you:biggrin:
 

Similar threads

  • · Replies 27 ·
Replies
27
Views
2K
Replies
4
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
1
Views
1K
Replies
7
Views
2K
Replies
7
Views
2K
Replies
8
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
4
Views
2K