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February 27th, 2010, 06:01 PM   #1
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First-Order Differential Equations

Need help with below two First-Order differential equation problems:

Problem# 1) xy (1+ xy^2) (dy/dx) = 1

Problem# 2) 2(dy/dx) = (y/x) - (x/y^2)

Appreciate your help so much.

Thank you.
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February 28th, 2010, 07:18 PM   #2
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# 1) xy(1 + xy)dy/dx = 1
This can be solved using a special function, but can you check it's typed correctly before I explain?

# 2) 2dy/dx = y/x - x/y
Let u = y + 3x, so that du/dx = 3ydy/dx + 6x = (3/2)(y/x - x) + 6x = (3/2)(u/x - 3x - x) + 6x = (3/2)u/x.
Hence x^(-3/2)du/dx - (3/2)x^(-5/2)u = 0.
Integrating gives x^(-3/2)u = C, i.e., u = Cx^(3/2), so y + 3x = Cx^(3/2).
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March 1st, 2010, 03:17 AM   #3
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Re: First-Order Differential Equations

Thanks a ton. Yes, I problem # 1) is typed correctly. Thanks for solving problem# 2)
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March 1st, 2010, 04:55 AM   #4
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Note that what I posted for #2 included a hidden assumption that x > 0. You can easily modify it to cover x < 0. I don't think it's possible for y to be differentiable at x = 0.

# 1) xy(1 + xy)dy/dx = 1

Let t = 1/x - 2, so that x = 1/(t + 2) and dx/dt = -1/(t + 2).
(y/(t + 2))(1 + y/(t + 2))dy/dt = dx/dt = -1/(t + 2).
y(y + t + 2)dy/dt = -1.

Let u = y + t, so that 2udu/dt = 2ydy/dt + 1.
(u + 2)(2udu/dt - 1) = -2.
2u(u + 2)du/dt - u = 0.
u = 0 or 2(u + 2/u)du/dt = 1.

Integrating the latter equation gives u + 4ln|u| = t + C.
From that, it's possible to get u in terms of t by using the Lambert W function.
More on this later.

The equation u = 0 leads to the solutions y = ?(2 - 1/x), where x < 0 or x > 1/2.
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March 1st, 2010, 10:29 AM   #5
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Re: First-Order Differential Equations

Thanks so much for your help! Just a quick question though... How did you know to use the substitution t = 1/x - 2 ? Do we just use guess and check or is there some sort of strategy?
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March 1st, 2010, 04:43 PM   #6
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I don't know of a simple strategy, but it's often worth trying various substitutions in a slightly generalized form, such as x = at^b, and then choosing values of a, b that seem to help. A slight change to the original equation may affect its difficulty substantially, but equations that are set as exercises or in examinations tend to be ones that can be done without excessive difficulty.
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