Questions - Differential equations

Questions / Answers

On 10/23/06, Anonymous wrote:
> Anonymous has left a new comment on your post "QUESTIONS -
> Lillian.Blackwell":
>
> Need some help!! Consider a second order differential equation of the form
> y" = f(y,y'). If we let v equal y', we obtain v' = f(y,v). This equation
> uses the variables x, y and v and hence is not the form of the first order
> equations. However, it is possible to eliminate the variable x by thinking
> of y as an independent variable; the chain rule gives dv/dx = (dv/dy)(dy/dx)
> = (dv/dy)v, and hence the original differential equation can be written as v
> (dv/dy) = f(y,v) Provided this equation can be solved, we obtain v as a
> function of y. A relation between y and x results from solving dy/dx = v(y)
> Again there are two arbitary constants in the final result. Use this
> technique to solve the following differential equation. y" + y(y')^3 = 0
> thanks
Posted by Anonymous to Maths,management,counselling at 9:29 AM
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Differential Equations
y" +y(y')^3 = 0
put
y'=dy/dx = u
y" = du/dx = (dy/dx)/(dy/du)=udu/dy
hence
udu/dy+yu^3=0
du/dy = -yu^2
-du/u^2=ydy
integrating
1/u = y^2/2 +k
u=dy/dx=1/(k+y^2/2)....you can integrate to get arc tan function ...