Higher Order Differential Equations, Solutions related

In summary: The equation is t^{-1}-2t^{-2}+2t^{-3}. Plugging that back into the differential equation gives you y'=-2t^{-1}+1.
  • #1
NCyellow
22
0

Homework Statement


Given the following differential equation
t^3y''' - t^2y'' + 2ty' - 2y = 0; t > 0
Find a solution that satisfies di fferential equation and the initial conditions
y(1) = 3; y'(1) = 2; y''(1) = 1

Homework Equations





The Attempt at a Solution


I tried plugging in 1 for y, y' and y'', but that gave me t^3y'''-t^2+2t-2=0, which doesn't give me anything... i looked in the book but still am positively stumped... Please help.
 
Physics news on Phys.org
  • #2
I don't understand why you would expect that to work! That would give you the value of y"' at x= 1 but no where else. The fact that they are equal to 1 at x= 1 tells you nothing about their values in general.

That equation simply says that
[tex]\frac{d^3y}{dt^3}= \frac{t^2- 2t+ 2}{t^3}= t^{-1}- 2t^{-2}+ 2t^{-3}[/tex]

Integrate three times.
 
  • #3
I'm sorry, how did you get that equation?
 
  • #4
Are you familiar with constant coefficient ode's?

If yes, then make the substitution x=ln(t), and find u(x)=y(e^x).
(Remember that the derivatives of y wrt to t are different, and use the chain rule)
 
  • #5
This type of DE is called an Euler or Euler-LaGrange equation. Look for a solution of the form y = tn. When you plug that into the DE you will get an equation that you can solve for n. It is similar to what happens for constant coefficient DE's. You can read about them, for example at:

http://tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx

In particular, that location shows you what happens if you get a repeated root for n.
 
  • #6
HallsofIvy said:
That equation simply says that
[tex]\frac{d^3y}{dt^3}= \frac{t^2- 2t+ 2}{t^3}= t^{-1}- 2t^{-2}+ 2t^{-3}[/tex]

Integrate three times.

NCyellow said:
I'm sorry, how did you get that equation?

Sometimes we all type before we look closely. I don't think Halls really meant that.
 

What are higher order differential equations?

Higher order differential equations are a type of differential equation that involves at least one derivative of a function with respect to an independent variable. They are generally more complex than first-order differential equations and require a higher level of mathematical understanding to solve.

How are higher order differential equations solved?

Higher order differential equations can be solved using a variety of methods, such as separation of variables, substitution, and series solutions. The specific method used will depend on the type and complexity of the equation.

What is the purpose of solving higher order differential equations?

Solving higher order differential equations allows us to model and understand various physical phenomena in fields such as physics, engineering, and economics. They also have numerous applications in real-world problems, such as predicting the behavior of complex systems.

Can higher order differential equations have multiple solutions?

Yes, higher order differential equations can have multiple solutions. In fact, in some cases, an equation may have an infinite number of solutions. This can occur when the equation is non-linear or has multiple initial conditions.

What are boundary value problems in the context of higher order differential equations?

Boundary value problems refer to a type of problem in which the solution of a differential equation is determined by specifying the values of the dependent variable at certain points, rather than just the initial conditions. They typically involve solving a higher order differential equation subject to certain boundary conditions.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
515
  • Calculus and Beyond Homework Help
Replies
2
Views
221
  • Calculus and Beyond Homework Help
Replies
2
Views
669
  • Calculus and Beyond Homework Help
Replies
6
Views
793
  • Calculus and Beyond Homework Help
Replies
7
Views
132
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
641
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top