Solution of the nonlinear 2nd order differential equation

In summary, the conversation is about solving a non-homogeneous second order differential equation with boundary conditions. The equation is given in terms of v, which is equal to the derivative of y. The suggested method is to use the cubic formula to find the roots and then integrate the solutions for v. However, the person asking for help expresses difficulty in applying this method due to its complexity.
  • #1
younginmoon
6
0

Homework Statement



I'd like to solve the following non-homogeneous second order differential equation and may
I ask smart scholars out there to help me with this?


y"(1-1.5(y')^2)=Cx^n, (^ denotes "to the power of")

where C and n are constants, and the boundary conditions are:
y=0 at x=0,
y'=0 at x=L/2 (L is between 100 and 200).

Thanks.

Homework Equations





The Attempt at a Solution


Indtroducing v=y', the equation becomes
v'(1.0-1.5v^2)=Cx^n
Integration of the above equation provides
(v-0.5v^3)=nCx^(N+1)-const.
Employing v=0 at x=L/2, const=nC(L/2)^(n+1), and the equation becomes
v-0.5v^3=nCx^(n+1)+nC(L/2)^(n+1)
I can't go any further.
 
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  • #2
I would suggest finding the roots of the cubic equation. It's already in depressed form ([tex]x^3 + bx + c = 0[/tex]), so it should be easy to solve with the cubic formula. Then all you need to do is integrate the three solutions for v. BTW, you made a mistake integrating the RHS.
 
  • #3
Thank you for the reply.
However, the only method I know to solve this kind of depressed cubic equation is the so-called Cardino formula that becomes a monster when you write it down by following the formula, and it's impossible to integrate. Just try to spread out the formula and apply each term, then you'll see the complexity of solution.
Anyway, thanks !
 

Related to Solution of the nonlinear 2nd order differential equation

1. What is the difference between a linear and nonlinear 2nd order differential equation?

A linear 2nd order differential equation can be written in the form of y'' + p(x)y' + q(x)y = g(x), where p(x) and q(x) are functions of x. On the other hand, a nonlinear 2nd order differential equation cannot be written in this form, as it involves a nonlinear relationship between the dependent variable and its derivatives.

2. How do you solve a nonlinear 2nd order differential equation?

There is no general method for solving a nonlinear 2nd order differential equation. However, there are some specific techniques for solving certain types of nonlinear equations, such as the substitution method, the power series method, or the method of undetermined coefficients.

3. Can a nonlinear 2nd order differential equation have multiple solutions?

Yes, a nonlinear 2nd order differential equation can have multiple solutions. Unlike linear equations, which have a unique solution, a nonlinear equation can have an infinite number of solutions that satisfy the given initial conditions.

4. What is the significance of solving a nonlinear 2nd order differential equation?

The solution of a nonlinear 2nd order differential equation can provide important insights into the behavior of a physical system. It can help predict the future behavior of the system and understand the underlying relationships between variables.

5. Are there any real-world applications of nonlinear 2nd order differential equations?

Yes, nonlinear 2nd order differential equations are widely used in various fields of science, such as physics, engineering, biology, and economics. They are used to model complex systems and phenomena, such as population growth, chemical reactions, and mechanical vibrations.

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