1st order ODE - I can't tell where to start

**Lancet** · June 21st 2011, 10:20 AM

I've got an ODE that I'm not even sure how to start with:

$y' - 4x^2y = xy*ln(y^{2x})$

As far as I can tell, it's not homogeneous, not seperable, not linear, and not exact.

I'm not sure what approach to take. Can someone help me figure out where to start?

**chisigma** · June 21st 2011, 11:29 AM

If You consider that $\ln y^{2x} = 2 x \ln y$ the ODE becomes separable...

Kind regards

$\chi$ $\sigma$

**Lancet** · June 21st 2011, 11:53 AM

Okay, if I do that, I can factor out the $2x^2$ , leaving the y terms grouped.

$y' - 2x^2(2y - yln(y)) = 0$

Is this supposed to be seperable? The seperable equations I've worked with were all multiplation/division at the top most levels though, whereas in this case, we have subtraction going on.

What you reminded me of definitely helped, but I'm still uncertain how to work with this.

**chisigma** · June 21st 2011, 12:16 PM

All right!... so we can separate the variables obtaining...

$\frac{dy}{y\ (2-\ln y)} = 2 x^{2}\ dx$ (1)

The next step now is to integrate both terms of (1). The integral in the x is not a problem... regarding the integral...

$\int \frac{dy}{y\ (2-\ln y)}$ (2)

... we can see an important detail: $\frac{1}{y}$ is the negate of derivative of $2 - \ln y$ ... what does it suggest that?...

Kind regards

$\chi$ $\sigma$

**Lancet** · June 21st 2011, 01:00 PM

Oh! Okay, I think I see. This is the first time I've worked with a seperable equation that had addition/subtraction top terms.

So, let me run this by you:

$2x^2 dx = \frac{1}{y(2 - ln(y))} dy$

$4x = -ln|2 - ln(y)| + c$

$e^{4x} = e^{-ln|2 - ln(y)| + c}$

$e^{4x} = |2 - ln(y)|^{-1}c$

$|2 - ln(y)|^{-1} = ce^{4x}$

$e^{|2 - ln(y)|} = e^{ce^{-4x}}$

$\frac{e^2}{y} = e^{ce^{-4x}}$

$e^2 = ye^{ce^{-4x}}$

$y = e^{2 - ce^{-4x}}$

How does that look?

**Ackbeet** · June 21st 2011, 06:44 PM

Originally Posted by Lancet

Oh! Okay, I think I see. This is the first time I've worked with a seperable equation that had addition/subtraction top terms.

So, let me run this by you:

$2x^2 dx = \frac{1}{y(2 - ln(y))} dy$

$4x = -ln|2 - ln(y)| + c$

If you're integrating the RHS, you should follow the Golden Rule of Math: do unto one side what you would do to the other.

$e^{4x} = e^{-ln|2 - ln(y)| + c}$

$e^{4x} = |2 - ln(y)|^{-1}c$

$|2 - ln(y)|^{-1} = ce^{4x}$

$e^{|2 - ln(y)|} = e^{ce^{-4x}}$

$\frac{e^2}{y} = e^{ce^{-4x}}$

$e^2 = ye^{ce^{-4x}}$

$y = e^{2 - ce^{-4x}}$

How does that look?

**Lancet** · June 21st 2011, 07:06 PM

Originally Posted by Ackbeet

If you're integrating the RHS, you should follow the Golden Rule of Math: do unto one side what you would do to the other.

Whoops! Okay, try this:

$2x^2 dx = \frac{1}{y(2 - ln(y))} dy$

$\frac{2}{3}x^3 = -ln|2 - ln(y)| + c$

$e^{\frac{2}{3}x^3} = e^{-ln|2 - ln(y)| + c}$

$e^{\frac{2}{3}x^3} = |2 - ln(y)|^{-1}c$

$|2 - ln(y)|^{-1} = ce^{\frac{2}{3}x^3}$

$e^{|2 - ln(y)|} = e^{ce^{-\frac{2}{3}x^3}}$

$\frac{e^2}{y} = e^{ce^{-\frac{2}{3}x^3}}$

$e^2 = ye^{ce^{-\frac{2}{3}x^3}}$

$y = e^{2 - ce^{-\frac{2}{3}x^3}}$

How is that now?

**Ackbeet** · June 21st 2011, 07:14 PM

Originally Posted by Lancet

Whoops! Okay, try this:

$2x^2 dx = \frac{1}{y(2 - ln(y))} dy$

$\frac{2}{3}x^3 = -ln|2 - ln(y)| + c$

$e^{\frac{2}{3}x^3} = e^{-ln|2 - ln(y)| + c}$

$e^{\frac{2}{3}x^3} = |2 - ln(y)|^{-1}c$

$|2 - ln(y)|^{-1} = ce^{\frac{2}{3}x^3}$

$e^{|2 - ln(y)|} = e^{ce^{-\frac{2}{3}x^3}}$

You can do that, but I'm not sure why you'd want to. Why not just invert the fraction?

$\frac{e^2}{y} = e^{ce^{-\frac{2}{3}x^3}}$

$e^2 = ye^{ce^{-\frac{2}{3}x^3}}$

$y = e^{2 - ce^{-\frac{2}{3}x^3}}$

How is that now?

**Lancet** · June 21st 2011, 07:26 PM

Originally Posted by Ackbeet

You can do that, but I'm not sure why you'd want to. Why not just invert the fraction?

To me, it made it neater to deal with the e and ln.

Is the final answer correct at this point? (Wolfram Alpha shows something a bit different, so I can't use that reliably to check.)

**Ackbeet** · June 22nd 2011, 02:32 AM

Originally Posted by Lancet

To me, it made it neater to deal with the e and ln.

Is the final answer correct at this point? (Wolfram Alpha shows something a bit different, so I can't use that reliably to check.)

Ah, but this is the beauty of DE's: it's quite straight-forward, usually, to check your own answer. Just plug it into the original DE and see if you get equality. What do you get?

Incidentally, there are two things you should always do when solving a DE. One is to check your answer by plugging back into the DE. The other is to determine the intervals of validity of the solution.

**Lancet** · June 22nd 2011, 03:05 AM

Originally Posted by Ackbeet

Ah, but this is the beauty of DE's: it's quite straight-forward, usually, to check your own answer. Just plug it into the original DE and see if you get equality. What do you get?

Incidentally, there are two things you should always do when solving a DE. One is to check your answer by plugging back into the DE. The other is to determine the intervals of validity of the solution.

Okay, well if I plug in and put it through WolframAlpha, the result is decidedly not zero, so there is still a mistake somewhere.

**Ackbeet** · June 22nd 2011, 03:11 AM

Why not go back to Post # 8, and try the different route suggested there?

**Lancet** · June 22nd 2011, 03:23 AM

Originally Posted by Ackbeet

Why not go back to Post # 8, and try the different route suggested there?

Okay, well first of all, you never said I made a mistake - it appeared as though I simply took a different approach to what you would have done.

That being said, would inverting the fraction not result in a 1/dx and 1/dy condition as well?

**Ackbeet** · June 22nd 2011, 05:13 AM

Originally Posted by Lancet

Okay, well first of all, you never said I made a mistake - it appeared as though I simply took a different approach to what you would have done.

Quite right. I didn't say you made a mistake. If you did, it might have something to do with a possibly unjustified elimination of absolute value signs. I think that, if anything, there might be some sign errors in your answer. Mathematica gives the answer as

$\{\{y[x]\to e^{-2+e^{\frac{2x^{3}}{3}+\frac{C[1]}{3}}}\}\},$

which is equivalent to

$y=e^{-2+Ce^{2x^{3}/3}}.$

Comparing this with your answer, you can see that there are sign differences in the exponential.

That being said, would inverting the fraction not result in a 1/dx and 1/dy condition as well?

Why? You've already integrated, so there shouldn't be any differentials in sight.

**Lancet** · June 22nd 2011, 01:45 PM

Originally Posted by Ackbeet

Comparing this with your answer, you can see that there are sign differences in the exponential.

Is that an indication that I made a mistake or is that just another way of writing the same answer?

Because when I plug either what I got or what you posted into WolframAlpha as part of the original ODE, I don't get a result of zero with either one.

Originally Posted by Ackbeet

Why? You've already integrated, so there shouldn't be any differentials in sight.

You quoted a section with six lines and didn't specify which one you were referring to. I thought you meant the first, as it was the most obvious fraction, and the last line of the section you quoted was a case where I *had* inverted the fractions. I had no choice but to guess what you were referencing.

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