Solving d^2x/dt^2 + a/x = b: Approximations?

  • Thread starter Thread starter gabi.petrica
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on solving the second-order differential equation d²x/dt² + a/x = b, where a and b are constants. The solution involves multiplying by x' and integrating to derive the relationship between x and t, resulting in the equation (1/√2) ∫(dx/√(bx + c - a ln{x})) = t + const. This method provides a way to express time as a function of position without making approximations, although further clarification is needed for complete resolution.

PREREQUISITES
  • Understanding of differential equations, specifically second-order equations.
  • Familiarity with integration techniques and logarithmic functions.
  • Knowledge of physics concepts related to motion and forces.
  • Basic proficiency in manipulating algebraic expressions and constants.
NEXT STEPS
  • Study advanced techniques for solving second-order differential equations.
  • Explore the method of integrating factors in differential equations.
  • Learn about the applications of logarithmic integration in physics problems.
  • Investigate numerical methods for approximating solutions to complex differential equations.
USEFUL FOR

Students and professionals in physics, mathematicians, and engineers who are dealing with differential equations in motion analysis and seeking exact solutions without approximations.

gabi.petrica
Messages
3
Reaction score
0
I was trying to solve a physics problem which led me to an equation of the form:
d^2x/dt^2 + a/ x = b; Can this be solved without any aproximations beeing made?
 
Physics news on Phys.org
Depends.

Assuming a,b are constants, multiply by [tex]x^{\prime}[/tex], to get

[tex]x^{\prime} x^{\prime \prime} + a \frac{x^{\prime}}{x} = b x^{\prime}[/tex]

now integrate to get

[tex]\frac{x^{\prime}^2}{2} + a \ln{x} = bx + c[/tex]

(c is a constant). Rearrange to get t as a function of x, i.e.,

[tex]\frac{1}{\sqrt{2}} \int{\frac{dx}{\sqrt{bx + c - a \ln{x}}}} = t + const.[/tex]

Other than that, I'm not sure.
 

Similar threads

  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K