solve for y: $xy'=y^2-2y$
$$xy'=y^2-2y$$ $$y(1) = 1, x >= 0$$
I get $C = 0$ which gives:
$$left| frac{y}{y-2}right|=x^2$$ When solving for $y$ it gives $$frac{2x^2}{1+x^2}$$, however the correct answer should be $$frac{2}{1+x^2}$$.
Is $$left| frac{y}{y-2}right|=x^2$$ correct? And if it is, how to continue from there?
differential-equations
asked May 19 '14 at 11:18
iveqy
721816
add a comment |
$$xy'=y^2-2y$$ $$y(1) = 1, x >= 0$$
I get $C = 0$ which gives:
$$left| frac{y}{y-2}right|=x^2$$ When solving for $y$ it gives $$frac{2x^2}{1+x^2}$$, however the correct answer should be $$frac{2}{1+x^2}$$.
Is $$left| frac{y}{y-2}right|=x^2$$ correct? And if it is, how to continue from there?
differential-equations
asked May 19 '14 at 11:18
iveqy
721816
add a comment |
$$xy'=y^2-2y$$ $$y(1) = 1, x >= 0$$
I get $C = 0$ which gives:
$$left| frac{y}{y-2}right|=x^2$$ When solving for $y$ it gives $$frac{2x^2}{1+x^2}$$, however the correct answer should be $$frac{2}{1+x^2}$$.
Is $$left| frac{y}{y-2}right|=x^2$$ correct? And if it is, how to continue from there?
differential-equations
asked May 19 '14 at 11:18
iveqy
721816
$$xy'=y^2-2y$$ $$y(1) = 1, x >= 0$$
I get $C = 0$ which gives:
$$left| frac{y}{y-2}right|=x^2$$ When solving for $y$ it gives $$frac{2x^2}{1+x^2}$$, however the correct answer should be $$frac{2}{1+x^2}$$.
Is $$left| frac{y}{y-2}right|=x^2$$ correct? And if it is, how to continue from there?
differential-equations
differential-equations
asked May 19 '14 at 11:18
iveqy
721816
asked May 19 '14 at 11:18
iveqy
721816
asked May 19 '14 at 11:18
iveqy
721816
asked May 19 '14 at 11:18
iveqy
721816
asked May 19 '14 at 11:18
iveqy
721816
721816
add a comment |
add a comment |
1 Answer
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(2018-11-25) "Amusing" revenge downvote, four years later.
Instead, try $$left| frac{y-2}{y}right|=x^2.$$
answered May 19 '14 at 11:28
Did
246k23220454
add a comment |
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1 Answer
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active
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(2018-11-25) "Amusing" revenge downvote, four years later.
Instead, try $$left| frac{y-2}{y}right|=x^2.$$
answered May 19 '14 at 11:28
Did
246k23220454
add a comment |
(2018-11-25) "Amusing" revenge downvote, four years later.
Instead, try $$left| frac{y-2}{y}right|=x^2.$$
answered May 19 '14 at 11:28
Did
246k23220454
add a comment |
(2018-11-25) "Amusing" revenge downvote, four years later.
Instead, try $$left| frac{y-2}{y}right|=x^2.$$
answered May 19 '14 at 11:28
Did
246k23220454
(2018-11-25) "Amusing" revenge downvote, four years later.
Instead, try $$left| frac{y-2}{y}right|=x^2.$$
answered May 19 '14 at 11:28
Did
246k23220454
edited Nov 25 at 17:55
answered May 19 '14 at 11:28
Did
246k23220454
answered May 19 '14 at 11:28
Did
246k23220454
answered May 19 '14 at 11:28
Did
246k23220454
246k23220454
add a comment |
add a comment |
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