Help to solve 4th order ode

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In summary, Kevin is trying to solve a fourth-order partial differential equation with given boundary conditions. The solution involves using the Fourier transform and finding the positive roots of a specific equation. Kevin is seeking help in understanding how this term is derived and also asks about using the Fourier transform to solve the equation.
  • #1
jimmygriffey
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Hi:

I try to solve this PDE as below:

y''''-ky''+1=0

With boundary conditions of x=0.y=y''=0 and x=-h,y=y'=0.
k and h are constant.
The solution look like from the Fourier transforms.
There is one term in the solution which is :

beta_n is the positive roots of the equation of beta_n=tan(beta_n). n is from 1 to infinity.

I have no idea how this term come from. If someone can help me, I really appreciate it!

Kevin
 
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  • #2
First of all your ODE has the following form

z''-kz+1=0, where z=y''

The general solution to z is obtained by the wery standard method which lead to ODE

y''= exp(k^(1/2)*x)*C1+exp(-k^(1/2)*x)*C2+1/k

and after double integration

y = C1/k*exp(k^(1/2)*x)+C2/k*exp(-k^(1/2)*x)+1/2/k*x^2+C3*x+C4

what is the general solution to your ODE. Substitution of your boundary conditions gives you 4 algebraic equations for unknowns C1,C2,C3 and C4.
 
  • #3
Almost the same thing: the characteristic equation for the differential equation y""- ky"+ 1= 0, which is the same as y""- ky"= -1, is [itex]r^4- kr^2= r^2(r^2- k)= r^2(r- \sqrt{k})(r+ \sqrt{k})[/itex] which has 0 as a double root and roots [itex]\sqrt{k}[/itex] and [itex]-\sqrt{k}[/itex].

The general solution to the associated homogeneous equation is [itex]z= C_1x+ C_2+ C_3e^{x\sqrt{k}}+ C_4e^{-x\sqrt{k}}[/itex]. Since the right hand side is a constant, normally we would try a constant solution but since a constant and x already satisfy the homogeneous solution, we try [itex]z= Ax^2[/itex]. Then [itex]z'= 2Ax[/itex], [itex]z"= 2A[/itex] and [itex]z"'= z""= 0[/itex]. The equation becomes 0+ k(2A)= -1 so A= -1/(2k). The general soution to that equation is
[tex]z(x)= C_1x+ C_2+ C_3e^{x\sqrt{k}}+ C_4e^{-x\sqrt{k}}-\frac{1}{2k}x^2[/tex]
 
  • #4
Hi:

Thank you for response. If I want to use Fourier transform to solve this one, how can I find the boundary conditons for y''(-h)? I need to use inverse Laplace transform to convert the y function. Therefore, it will become much easier to use the Fourier transforms.

Kevin
 

Related to Help to solve 4th order ode

1. What is a 4th order ordinary differential equation (ODE)?

A 4th order ODE is a type of mathematical equation that involves a function and its derivatives, up to the 4th derivative. It is used to model a wide range of physical phenomena in science and engineering, such as motion, heat transfer, and electrical circuits.

2. How do you solve a 4th order ODE?

Solving a 4th order ODE involves finding the general solution, which is a function that satisfies the equation for all possible values of the independent variable. This can be done analytically using techniques such as separation of variables and variation of parameters, or numerically using computational methods such as Euler's method or Runge-Kutta methods.

3. What are the initial conditions for a 4th order ODE?

The initial conditions for a 4th order ODE are the values of the function and its first three derivatives at a specific point. These conditions are necessary to obtain a unique solution to the equation.

4. Can all 4th order ODEs be solved analytically?

No, not all 4th order ODEs have analytical solutions. In fact, most real-world problems require numerical methods to obtain an approximate solution. However, certain types of ODEs, such as linear homogeneous equations, can be solved analytically using well-known techniques.

5. What are some applications of 4th order ODEs in science?

4th order ODEs are widely used in many fields of science and engineering, including physics, biology, and chemistry. They can be used to model complex systems and phenomena, such as the motion of a pendulum, the behavior of a vibrating string, or the heat transfer in a rod.

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