Finding the General Solution for a Second Order Differential Equation

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SUMMARY

The discussion centers on the second order differential equation given by the expression \(\sqrt{1+(y')^2}=\frac{d}{dx}\left(y\frac{y'}{\sqrt{1+(y')^2}}\right)\). A specific solution is identified as \(y(x)=\cosh(x+C_{1})\), but the poster seeks a more general solution that accounts for the two degrees of freedom inherent in second order ordinary differential equations (ODEs). The inquiry emphasizes the need for a comprehensive approach to finding the general solution.

PREREQUISITES
  • Understanding of second order ordinary differential equations (ODEs)
  • Familiarity with hyperbolic functions, specifically \(\cosh\)
  • Knowledge of differential calculus and derivatives
  • Experience with solving differential equations using analytical methods
NEXT STEPS
  • Research methods for solving second order ODEs, focusing on the method of characteristics
  • Explore the concept of general solutions in the context of differential equations
  • Study the application of boundary conditions to determine specific solutions
  • Investigate the use of software tools like Mathematica or MATLAB for solving complex differential equations
USEFUL FOR

Mathematicians, engineering students, and researchers in applied mathematics who are working with differential equations and seeking to deepen their understanding of second order ODEs and their solutions.

Daniel D
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Hi,

I came across the following differential equation:

[tex]\sqrt{1+(y')^2}=\frac{d}{dx}\left(y\frac{y'}{\sqrt{1+(y')^2}}\right)[/tex]

I found possible solutions: [tex]y\left(x\right)=cosh(x+C_{1})[/tex].
However, this is a second order ODE so there exist a more general solution, with 2 freedom degrees.

Can anyone find it?

Thank you.
 
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Daniel D said:
Hi,

I came across the following differential equation:

[tex]\sqrt{1+(y')^2}=\frac{d}{dx}\left(y\frac{y'}{\sqrt{1+(y')^2}}\right)[/tex]

I found possible solutions: [tex]y\left(x\right)=cosh(x+C_{1})[/tex].
However, this is a second order ODE so there exist a more general solution, with 2 freedom degrees.

Can anyone find it?

Thank you.

What do you mean "can anyone find it"? Please show us how you have tried to find it so far. This is the PF, not Yahoo answers.
 

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