Interesting 2nd order non-linear DE

**KrayzBlu** · February 2nd 2011, 11:32 AM

Hello,

I have recently come across an interesting second order non-linear homogeneous DE whose form I cannot seem to find a general solution to. It loos like such:

$\dfrac{d^2}{dx^2}f(x) + B(x)*\dfrac{d}{dx}f(x) + f(x)*\dfrac{d}{dx}B(x) = 0$

Where B(x) is a arbitrary function. Has anyone seen this before or have any hints?

Thanks

**TheEmptySet** · February 2nd 2011, 11:36 AM

Originally Posted by KrayzBlu

Hello,

I have recently come across an interesting second order non-linear homogeneous DE whose form I cannot seem to find a general solution to. It loos like such:

$\dfrac{d^2}{dx^2}f(x) + B(x)*\dfrac{d}{dx}f(x) + f(x)*\dfrac{d}{dx}B(x) = 0$

Where B(x) is a arbitrary function. Has anyone seen this before or have any hints?

Thanks

Notice that

$\displaystyle \frac{d}{dx}\left[B(x)f(x)\right]=B(x)\frac{df}{dx}+f(x)\frac{dB}{dx}$

This gives the equation as

$\displaystyle \frac{d}{dx}\left[ \frac{df}{dx}+B(x)f(x)\right]=0$

Now just integrate both sides and you have a first order equation that can be solved via an integrating factor.

**Ackbeet** · February 2nd 2011, 11:41 AM

It's nonlinear?

**KrayzBlu** · February 2nd 2011, 01:15 PM

You're right! I'm embarrassed that I missed that

.

Thank you, sir.

(Although interestingly, alpha couldn't solve it either f''(x) + b(x)*f'(x) + b'(x)*f(x) = 0 - Wolfram|Alpha)

As to Ackbeet, from what I understand, and from what alpha seems to confirm, if B(x) could be non-linear, then the equation is non-linear. Please correct me if I'm wrong! (I could never quite get this straight in school)

**TheEmptySet** · February 2nd 2011, 01:27 PM

Originally Posted by KrayzBlu

You're right! I'm embarrassed that I missed that

.

Thank you, sir.

(Although interestingly, alpha couldn't solve it either f''(x) + b(x)*f'(x) + b'(x)*f(x) = 0 - Wolfram|Alpha)

As to Ackbeet, from what I understand, and from what alpha seems to confirm, if B(x) could be non-linear, then the equation is non-linear. Please correct me if I'm wrong! (I could never quite get this straight in school)

I would not classify it as nonlinear. The equation just has non constant coefficients. It would only be nonlinear if $B(x)$ somehow depened on $f(x)$ as an input.

For example is $B(x)=\sin(x) \implies B'(x)=\cos(x)$

Then the equation would be

$f''(x)+\sin(x)f'(x)+cos(x)f(x)=0$

This is a 2nd order linear ODE

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