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Solve This Differential Equation: Dx/dy=3xy^2?

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Solve This Differential Equation: Dx/dy=3xy^2?

Postby Mamoon » Tue Sep 26, 2017 7:19 am

You originally wrote Dx/Dy instead of dy/Dx, is this correct?

I will show you for both!

If Dx/dy=3xy^2

Then Dx/3x=y^2dy

So integrate both, and you get 1/3ln3x = y^3/3

So y^3 = ln3x

So y= (cube root)ln3x + c

If dy/Dx=3xy^2

Then dy/y^2=3xdx

So integrate both and you get -1/y = 3/2x^2

So, y=-2/(3x^2) + c

Hope this helps! :)
Mamoon
 
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Solve This Differential Equation: Dx/dy=3xy^2?

Postby Aisley » Wed Sep 27, 2017 12:44 pm

Going from -1/y=3/2x^2

Multiply both sides by 2x^2 getting -2x^2/y = 3

Multiply both sides by y to get -2x^2=3y, divide both sides by 3 to solve for y to get

y=-2(x^2) / 3
Aisley
 
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Solve This Differential Equation: Dx/dy=3xy^2?

Postby ozzi » Thu Sep 28, 2017 1:42 am

dx/dy = 3xy^2

(x^-1)x = (3y^2)dy

lnx = y^3 + c

y^3 = lnx - c

y = (lnx - c)^(1/3)

===
ozzi
 
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Solve This Differential Equation: Dx/dy=3xy^2?

Postby Norwin » Thu Sep 28, 2017 2:22 pm

I doubt very so much that a closed type resolution exists. It might, but i'd be amazed. However it lends itself fantastically for numerical fixing. 1) y' is constantly +ve so the operate is certainly not decreasing 2) as x --> ±?, y' --> zero So y(x) may just seem similar to the erf(x) perform. *** good truely I feel the reply is y(x) = 0 identically. If y'(zero) = zero, then y certainly not will get round to relocating from y(zero) = zero. **** For other preliminary conditions, y(x) will resemble the error function, but it'll be contained between consecutive values n?. That is for the reason that each time y = n?, y' = zero, and as a result y=n? will constantly be an aymptote to the operate.
Norwin
 
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Solve This Differential Equation: Dx/dy=3xy^2?

Postby Athdara » Sat Sep 30, 2017 10:48 pm

3y^2 dy=dx/x

y^3=ln x + c
Athdara
 
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