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

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

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

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

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?

2. ## Re: 1st order ODE - I can't tell where to start

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## Re: 1st order ODE - I can't tell where to start

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

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

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.

4. ## Re: 1st order ODE - I can't tell where to start

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. ## Re: 1st order ODE - I can't tell where to start

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:

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

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

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

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

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

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

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

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

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

How does that look?

6. ## Re: 1st order ODE - I can't tell where to start

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:

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

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

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

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

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

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

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

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

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

How does that look?

7. ## Re: 1st order ODE - I can't tell where to start

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:

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

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

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

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

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

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

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

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

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

How is that now?

8. ## Re: 1st order ODE - I can't tell where to start

Originally Posted by Lancet
Whoops! Okay, try this:

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

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

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

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

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

$\displaystyle e^{|2 - ln(y)|} = e^{ce^{-\frac{2}{3}x^3}}$
You can do that, but I'm not sure why you'd want to. Why not just invert the fraction?

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

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

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

How is that now?

9. ## Re: 1st order ODE - I can't tell where to start

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.)

10. ## Re: 1st order ODE - I can't tell where to start

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.

11. ## Re: 1st order ODE - I can't tell where to start

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.

12. ## Re: 1st order ODE - I can't tell where to start

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

13. ## Re: 1st order ODE - I can't tell where to start

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?

14. ## Re: 1st order ODE - I can't tell where to start

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

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

which is equivalent to

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

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.

15. ## Re: 1st order ODE - I can't tell where to start

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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