Solving ODE: 2y(1+x^2√y)dx + x(2+x^2√y)dy = 0 | Step-by-Step Guide

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The discussion focuses on solving the ordinary differential equation (ODE) given by 2y(1+x^2√y)dx + x(2+x^2√y)dy = 0. A substitution of x^2√y=u was attempted, but the differentiation proved complex. The correct differentiation was confirmed, leading to the suggestion to substitute for y and y' in the original equation to simplify it into a separable ODE. Another approach proposed involves grouping terms into perfect differentials, suggesting the use of u = xy and v = another expression to transform the problem into a linear ODE. The conversation emphasizes the importance of simplifying the equation for easier resolution.
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Homework Statement


2y(1+x^2\sqrt{y})dx + x(2+x^2\sqrt{y})dy = 0

The Attempt at a Solution


well, I substituted x^2√y=u but then when I tried to differentiate it I understood it would be so hard. Please check and see if I've differentiated it correctly:

√y = u/x^2 -> y = u^2.x^-4 -> dy/dx = 2u.u'x^(-4) - 4x^(-5).u^2
Is that correct? if yes, then I think I've just made the problem harder. how can I solve that ODE?
 
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It is correct. Substitute for y and y' in the original equation, simplify and arrange: it is a separable ODE.

ehild
 
You could try grouping the terms into perfect differentials. For example, 2 y dx + 2 x dy = 2 d(xy). The other two terms can be written similarly as d(something) . Then write u = xy and v = something and it is a simple linear ODE.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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