# 2nd O.D.E.

• December 21st 2008, 04:07 AM
tsal15
2nd O.D.E.
Hey all :)

How do I find an 'explicit formula' for a solution y(x), which also satisfies initial conditions y'(0) = 1 and y(0) = 0? - where the non-linear 2nd O.D.E is $\frac{d^2y}{dx^2} + 2x(\frac{dy}{dx})^2 = 0$

*_*_*_*_*_* MERRY CHRISTMAS*_*_*_*_*_*

• December 21st 2008, 04:29 AM
Mush
Quote:

Originally Posted by tsal15
Hey all :)

How do I find an 'explicit formula' for a solution y(x), which also satisfies initial conditions y'(0) = 1 and y(0) = 0? - where the non-linear 2nd O.D.E is $\frac{d^2y}{dx^2} + 2x(\frac{dy}{dx})^2 = 0$

*_*_*_*_*_* MERRY CHRISTMAS*_*_*_*_*_*

Let $u = \frac{dy}{dx}$

And hence $\frac{du}{dx} = \frac{d^2y}{dx^2}$

You can then rearrange to get:

$\frac{du}{u^2} = -2xdx$

Which is solveable by some fairly simple integration :), don't forget to sub back into y.

Explicit just means you write it in the form $y(x) = blah blah$

This works as a general rule btw. If there is no "y" term in your 2nd order ODE, then you can transform it into a first order ODE using substitution as shown above.
• December 21st 2008, 05:42 AM
tsal15
Quote:

Originally Posted by Mush
Let $u = \frac{dy}{dx}$

And hence $\frac{du}{dx} = \frac{d^2y}{dx^2}$

You can then rearrange to get:

$\frac{du}{u^2} = -2xdx$

Which is solveable by some fairly simple integration :), don't forget to sub back into y.

Explicit just means you write it in the form $y(x) = blah blah$

This works as a general rule btw. If there is no "y" term in your 2nd order ODE, then you can transform it into a first order ODE using substitution as shown above.

Great thanks, Mush.

I'll have a go at it again, and I'll get back to you :)

Thanks again :)
• December 25th 2008, 07:43 PM
tsal15
Quote:

Originally Posted by Mush
Let $u = \frac{dy}{dx}$

And hence $\frac{du}{dx} = \frac{d^2y}{dx^2}$

You can then rearrange to get:

$\frac{du}{u^2} = -2xdx$

Which is solveable by some fairly simple integration :), don't forget to sub back into y.

Explicit just means you write it in the form $y(x) = blah blah$

This works as a general rule btw. If there is no "y" term in your 2nd order ODE, then you can transform it into a first order ODE using substitution as shown above.

Hey, Mush.

Apparently I'm supposed to find the general equation first. How would I go about doing this?
• December 25th 2008, 08:07 PM
Jester
Quote:

Originally Posted by tsal15
Hey all :)

How do I find an 'explicit formula' for a solution y(x), which also satisfies initial conditions y'(0) = 1 and y(0) = 0? - where the non-linear 2nd O.D.E is $\frac{d^2y}{dx^2} + 2x(\frac{dy}{dx})^2 = 0$

*_*_*_*_*_* MERRY CHRISTMAS*_*_*_*_*_*

If you let $y' = u$, the ODE separates giving

$\frac{-u'}{u^2} = x$.

Integrating, applying the IC and integrating again with the second IC gives the solution

$y = tan^{-1} x$.
• December 25th 2008, 08:31 PM
tsal15
Quote:

Originally Posted by danny arrigo
If you let $y' = u$, the ODE separates giving

$\frac{-u'}{u^2} = x$.

Integrating, applying the IC and integrating again with the second IC gives the solution

$y = tan^{-1} x$.

I presume that IC = Initial Condition ?

Also whats the general solution? - is it \frac{u'}{u^2} = x , if not, how do find the general solution ? :)

Thanks Danny,

tsal15
• December 25th 2008, 09:08 PM
Jester
Quote:

Originally Posted by tsal15
I presume that IC = Initial Condition ?

Also whats the general solution? - is it \frac{u'}{u^2} = x , if not, how do find the general solution ? :)

Thanks Danny,

tsal15

Yes, IC is the initial condition. Let me provide some more details

$-\frac{u'}{u^2} = 2x \;\;\; \Rightarrow\;\;\;-\int \frac{u'}{u^2} = \int 2x +c_1$

so

$\frac{1}{u} = x^2+c_1 \;\;\; \Rightarrow\;\;(with \;y'(0)\;=\;u(0)=1)\;\;\frac{1}{u} = x^2+1$

so that

$u = y' = \frac{1}{x^2+1}$

Integrating gives

$y = tan^{-1} x + c_2$

and with the IC $y(0) = 0$ gives our result.
• December 25th 2008, 10:08 PM
mr fantastic
Quote:

Originally Posted by tsal15
Hey, Mush.

Apparently I'm supposed to find the general equation first. How would I go about doing this?

You would surely use the given boundary conditions as you go along.

It's ridiculous to carry an arbitrary constant into the second integration when it's possible to get its value. Not the least reason being that the technique required for doing the second integration will depend on what the value of that first arbitrary constant is ....
• January 12th 2009, 07:04 PM
tsal15
Finding the general solution of the 2ODE, was the process we have been discussing so far?