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How to Integrate dy/dx= 2xy?

Is it y=xy^2 ?
Reply 1
Avatar for Notnek
Notnek
21
Original post by Airess3
Is it y=xy^2 ?

If you differentiate that (implicitly) then you'll find that it's incorrect.

You need to 'separate the variables'. Do you know how to do this? If not, can you say where you got this question from and what level you're at?
Original post by notnek
If you differentiate that (implicitly) then you'll find that it's incorrect.

You need to 'separate the variables'. Do you know how to do this? If not, can you say where you got this question from and what level you're at?


So 2 dy/dx xy? It still comes down to the same answer. (2xy^2)/2 so xy^2
Reply 3
Avatar for Notnek
Notnek
21
Original post by Airess3
So 2 dy/dx xy?

Where has this come from?

I don't understand you post.

EDIT : I think you're trying to differentiate. You should be using the product rule.
(edited 8 years ago)
Original post by notnek
Where has this come from?

I don't understand you post.

EDIT : I think you're trying to differentiate. You should be using the product rule.


No, I'm trying to integrate, not use the product rule because that's differentiation.
since you integrate with respect to the certain variable, try putting the x's and dx's on one side and the y's and dy's on one side. The 2's location in this doesn't matter because it's a constant
Original post by Airess3
Is it y=xy^2 ?
If dy/dx = f(x)/g(y), then

∫g(y) dy=∫f(x) dx\int g(y)\,dy = \int f(x)\,dx (you'll have to consider the arbitrary constants after integrating).
Reply 7
Avatar for hlyon200
hlyon200
8
dy/dx=2xy

(1/y)dy=(2x)dx

Edit: solve the rest yourself
(edited 8 years ago)
Reply 8
Avatar for Ayman!
Ayman!
15
Original post by hlyon200
dy/dx=2xy

(1/y)dy=(2x)dx



It's against the spirit of this forum to post a full solution. I'd advice you to remove your solution before the mods remove it
(edited 8 years ago)
Reply 9
Avatar for hlyon200
hlyon200
8
Original post by aymanzayedmannan
It's against the spirit of this forum to post a full solution. I'd advice you to remove your solution before the mods remove it


Done. I didn't realise!!! My previous working is now in your quote however
Original post by DFranklin
If dy/dx = f(x)/g(y), then

∫g(y) dy=∫f(x) dx\int g(y)\,dy = \int f(x)\,dx (you'll have to consider the arbitrary constants after integrating).


So I got 1/2y -x^2 . After I moved the y to the right hand side. Is this right?
(edited 8 years ago)
Original post by Airess3
So I got 1/2y -x^2 . After I moved the y to the right hand side. Is this right?


Try this - split your original equation so you have the y and dy on one side and the 2x and dx on the other and integrate both sides.

When you collect the same variable and d(variable) on one side, you can integrate them as normal, so by splitting it into all y's on one side and all x's on the other, you pretty much make it two normal integrations to get what y and x are.

integral (1/y)dy = integral 2x dx
(edited 8 years ago)
Reply 12
Avatar for Notnek
Notnek
21
Original post by Airess3
So I got 1/2y -x^2 . After I moved the y to the right hand side. Is this right?

I'm not sure what you've done here. Please try to explain your working each time you post.

Have you done 'separation of variables before'?

Also, what level are you at and where did you get the question from?
Original post by not you
Try this - split your original equation so you have the y and dy on one side and the 2x and dx on the other and integrate both sides.

When you collect the same variable and d(variable) on one side, you can integrate them as normal, so by splitting it into all y's on one side and all x's on the other, you pretty much make it two normal integrations to get what y and x are.

integral (1/y)dy = integral 2x dx


So all I'm left with is x^2?
Original post by Airess3
So all I'm left with is x^2?


Not quite, that's just the x side (also remember your +c). You still have to integrate 1/y, and when you integrate 1/variable you get ln(variable) so you'll get ln(y) = x^2 + c

Once you get that, you'll want to get rid of that ln, so go back to your log rules and see how you could get y=
(edited 8 years ago)
Reply 15
Avatar for Ayman!
Ayman!
15
Suppose you have the following differential equation

Unparseable latex formula:

\[\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{x^{n}}{y^{n}}



We now use a process called separating the variables to obtain a general solution and make integration easier. In layman's terms, I could say that I'm treating dy/dx as a fraction and taking the x's and the dx's on one side and the y's and the dy's on the other (it is an incorrect analogy mathematically, but helpful for visualisation).

⇒yn⋅dy=xn⋅dx\Rightarrow y^{n}\cdot dy = x^{n}\cdot dx

Finally, we integrate both sides to obtain a solution

Unparseable latex formula:

\Rightarrow \int y^{n} dy = \int x^{n}dx\]



I hope my example is clear.
Original post by Airess3
So all I'm left with is x^2?
Do you know what ∫1y dy\displaystyle \int \frac{1}{y}\,dy is? (The normal rule ∫yk dy=yk+1k+1\int y^k \,dy = \frac{y^{k+1}}{k+1} doesn't work when k = -1).

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