differential equation y(x)''=f(y(x))

Holali

Hi,
simple quetion, as you can see in the title.
How can I solve differential equation y(x)''=f(y(x))
I know I can write first derivative like dy/dx. But how can I write second derivative in such form?
If it would be y(x)'=y, then it can be writen dy/dx=y
=> (1/y)dy=(1)dx
=> I can integrate ln(y)+C=x
=> its clear from now on

I need some similar solution for situation when I have second derivative on the left side and some function consisting just y(x) (not x itself) on the right.
Thanks for helping

tiny-tim

Hi Holali!

Multiply both sides by y'

(or use the chain rule … y'' = dy'/dx = y' dy'/dy)

Holali

Hmm, chain rule.I found it at wikipedia,but cant understand it.
I understand equation y''=dy'/dx, but not y''=y' dy'/dy.
Could you show me some easy example, like y''=2y^2, or y''=y^2 -y ?

HallsofIvy

What tiny-time is referring to is often called "quadrature"- you'll see why in a moment. The crucial point is that "x", the independent variable does not appear explicitely in the equation y''= f(7).

If you let u= y', you can write
[tex]\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)= \frac{du}{dx}[/tex]
We can then say , using the chain rule,
[tex]\frac{du}{dx}= \frac{du}{dy}\frac{dy}{dx}= u\frac{du}{dy}= f(y)[/tex]
which is a separable equation:
[tex]u du= f(y)dy[/tex]
The left side is, of course, [itex](1/2)u^2[/itex], the reason for the name "quadrature". we have
[tex](1/2)u^2= \int f(x)dx[/tex]
so
[tex]\frac{dy}{dx}= u= \sqrt{2\int f(x)dx}[/tex]
[tex]y= \int^x \sqrt{2\int^u f(t)dt} du[/tex]

Of course, if f(y) was not a "nice" function to begin with, those integrals may be difficult to do!

Holali

""
udu=f(y)dy

The left side is, of course, (1/2)u2, the reason for the name "quadrature". we have

(1/2)u2=∫f(x)dx

""
DIdn't you change y for x in this part?

GarageDweller

Simple manipulation of left hand side of the equation will solve this

d(dy/dx)/dx= dy'/dy * dy/dx
Moving the 1/dy factor to the right hand side, we get

y' dy' = f(y) dy

Integrating this will give us

(1/2)y'^2=F(y) (F'(y)=f(y))

Rearranging, we get

dy/dx=[2F(y)]^1/2

And of course, the second order diff eq. is reduced to a simple ODE. Using the method of separation of variables, we obtain

x=integral{dy/[2F(y)]^1/2}

etpatati1

For the same general equation, how can i solve

y''(x)=a*sin(y(x)) ; a is a constant
I think it would be a numerical method.

tiny-tim

hi etpatati1! welcome to pf!
Multiply both sides by y'

(or use the chain rule … y'' = dy'/dx = y' dy'/dy)

JJacquelin

Thanks to the Jacobi am function, a closed form exists to express y(x).