Find the equations of the trajectories of y"+y^3=0

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The discussion focuses on solving the differential equation y'' + y³ = 0 to find the trajectories. The solution approach involves substituting y'' with p(y) and integrating to arrive at the equation v²/2 + y⁴/4 = C. However, the final goal is to express y as a function of the independent variable, which requires further manipulation of the derived equation. The correct next step is to solve the differential equation derived from the expression (1/2)(y')² = C - (1/4)y⁴.

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Homework Statement


Find the equations of the trajectories of y"+y^3=0.

Homework Equations


None.

The Attempt at a Solution


y"+p(y)=0
v(dv/dy)+p(y)=0
integrate
v^2/2+P(y)=C
so I got v^2/2+y^4/4=C. Is v^2/2+y^4/4=C the correct answer?
 
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Math10 said:

Homework Statement


Find the equations of the trajectories of y"+y^3=0.

Homework Equations


None.

The Attempt at a Solution


y"+p(y)=0
v(dv/dy)+p(y)=0
integrate
v^2/2+P(y)=C
so I got v^2/2+y^4/4=C. Is v^2/2+y^4/4=C the correct answer?

Substitute it into the DE and check to see if it works. You can do it just as easily as we can, and (as I have said before), doing your own checking first is a good habit to develop. You won't be able to ask for help when you write an exam!
 
Nope. You're looking for an expression where y is expressed as a function of x (or whatever your independent variable is). Right now, what you have is
$$\frac 12 (y')^2 = C - \frac 14 y^4.$$ You need to solve that differential equation now.
 

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