Non-linear 2nd ODE involving squares of derivatives

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SUMMARY

The discussion focuses on solving the non-linear second-order ordinary differential equation (ODE) given by y'' + (1/y)(y')² = 0. The key insight is to multiply through by y, transforming the equation into y y'' + (y')² = 0, which can be approached using the Product Rule. This leads to a separable differential equation. Additionally, by substituting y' with u, the equation can be simplified to a first-order equation for u, facilitating the solution process.

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  • Understanding of second-order ordinary differential equations (ODEs)
  • Familiarity with the Product Rule in calculus
  • Knowledge of separable differential equations
  • Basic concepts of substitution methods in differential equations
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  • Study the method of solving separable differential equations
  • Learn about the Product Rule and its applications in differential equations
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Students and educators in mathematics, particularly those focusing on differential equations, as well as researchers seeking to understand non-linear ODEs and their solution methods.

iqjump123
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Homework Statement



y''+(1/y)*(y')2=0

Homework Equations





The Attempt at a Solution



This is another problem I am having trouble with. I have done searches around the internet, but seen that all "non linear" ODE of second order involves a non linear form in a non differential term (like y''+xy^2=0, or something like that), instead of the DE term.

Punching through wolfram alpha gave a really simple straightforward answer, so I believe it shouldn't be too hard, as long as I get the general method to solve it.

thanks in advance.
 
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The question to ask might be, "Where have we seen something like this:

y'' + (1/y)*(y')2 = 0 ?"

We can see that y can't be zero, so let's multiply through by it to get

y y'' + (y')2 = 0 .

Now think about the Product Rule -- what is this the derivative of? You will then be led to a separable differential equation. (And I feel like I'm talking like a fortune cookie...)
 
A different hint, which can be applied to all second-order equations which do not contain the independent variable x explicitly: let be the independent variable y, and denote y'=u. Then

y''=u'=(du/dy)(dy/dx)=u (du/dy),

and you get a first-order equation for u.

ehild
 

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