Can anyone check if my answer for this ODE is correct?

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SUMMARY

The discussion centers on verifying the solution to the ordinary differential equation (ODE) given by the expression (x+2y-4)dx - (2x-4y)dy=0. The user's solution is ln[4(y-1)^2+(x-2)^2] + 2arctan((x-2)/(2y-2))=C, while the textbook's answer is ln[4(y-1)^2+(x-2)^2] - 2arctan((2y-2)/(x-2))=C. The discrepancy is attributed to the use of the identity tan(a) = 1/tan(π/2 - a), confirming that both solutions are equivalent except for a constant. Initial conditions were also discussed but not specified.

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ronaldor9
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Homework Statement


(x+2y-4)\,dx - (2x-4y)\,dy=0

The Attempt at a Solution


My answer,
\mathrm{ln}[4(y-1)^2+(x-2)^2] + 2\mathrm{arctan}\frac{x-2}{2y-2}=C
Textbook's answer,
\mathrm{ln}[4(y-1)^2+(x-2)^2] - 2\mathrm{arctan}\frac{2y-2}{x-2}=C
 
Last edited:
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Hi ronaldor9! :smile:

tan(a) = 1/tan(π/2 - a), so theyr'e the same, except for a constant :wink:

what were the initial conditions?
 
Hey, sorry i meant to place a constant instead of the zero.
Does that identity work for arctan?
 
ronaldor9 said:
Hey, sorry i meant to place a constant instead of the zero.
Does that identity work for arctan?

oh come on

if tan(a) = 1/tan(π/2 - a),

then putting T = tan(a) gives 1/T = tan(π/2 - a),

so arctanT = a = π/2 - arctan(1/T)
 
Thanks! tiny-tim
 
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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