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Help me to solve these differential equations
- Thread starter cloudyy
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HallsofIvy
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- Jan 27, 2012
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II \(\displaystyle \frac{dy}{dx}=\frac{-(x+1)}{y- 2}\) is the same as (y- 2)dy= -(x+1)dx.
Integrate!
However you will not get the result shown! It should be \(\displaystyle (y- 2)^2\), not \(\displaystyle (y-1)^2\).
III Do you know what a "direction field" is? At each (t, y) point draw a short line segment with slope 5- 2y. That does not depend upon "t" so, for example, at each point on the y= 0 axis, (t, 0), you will have short line with slope 5, pretty steep. At each point on the y= 1 line, (t, 1) you will have short line with slope 3, y= 2, slope 1, y= =3, -1, etc.
Integrate!
However you will not get the result shown! It should be \(\displaystyle (y- 2)^2\), not \(\displaystyle (y-1)^2\).
III Do you know what a "direction field" is? At each (t, y) point draw a short line segment with slope 5- 2y. That does not depend upon "t" so, for example, at each point on the y= 0 axis, (t, 0), you will have short line with slope 5, pretty steep. At each point on the y= 1 line, (t, 1) you will have short line with slope 3, y= 2, slope 1, y= =3, -1, etc.
I found Halls of Ivy’s answer to be correct but confusing. He is correct that the book has a typographical error, but Halls of Ivy misread the error of 3 for an error of 2. The book should have said
[MATH]\text {Show that } \dfrac{dy}{dx} = \dfrac{-(x + 1)}{y - 1} \iff (x + 1)^2 + (y - 1)^2 = c^2.[/MATH]
You will have to be a bit clever about constants of integration.
Moreover the “two families” part of the question is obscure. Try getting to y = f(x). The “two families will then become crystal clear.
EDIT: To be fair to Halls, his suggested first step
[MATH]\dfrac{dy}{dx} = \dfrac{-(x+1)}{(y-1)} \implies (y - 1)dy = -(x+1)dx[/MATH]
is spot on as a first step in showing
[MATH]\dfrac{dy}{dx} = \dfrac{-(x + 1)}{y - 1} \implies (x + 1)^2 + (y - 1)^2 = c^2.[/MATH]
[MATH]\text {Show that } \dfrac{dy}{dx} = \dfrac{-(x + 1)}{y - 1} \iff (x + 1)^2 + (y - 1)^2 = c^2.[/MATH]
You will have to be a bit clever about constants of integration.
Moreover the “two families” part of the question is obscure. Try getting to y = f(x). The “two families will then become crystal clear.
EDIT: To be fair to Halls, his suggested first step
[MATH]\dfrac{dy}{dx} = \dfrac{-(x+1)}{(y-1)} \implies (y - 1)dy = -(x+1)dx[/MATH]
is spot on as a first step in showing
[MATH]\dfrac{dy}{dx} = \dfrac{-(x + 1)}{y - 1} \implies (x + 1)^2 + (y - 1)^2 = c^2.[/MATH]
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