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course Mth 279
Diffeerential Equations Notes 110207`q001. Solve the equation y ' = t^(3/2) tan(y) for the initial condition y(pi) = 1.
y ' = t^(3/2) tan(y)
dy/tan(y) = t^(3/2) dt
Int(dy/tan(y))
= Int(cos(x)/sin(x) dy), using u sub, u = sin(y), du = cos(y) dy
Int(1/u du) = ln|u| = ln|sin(y)|
Int(t^(3/2) dt) = (2/5)t^(5/2) + C
ln|sin(y)| = (2/5)t^(5/2) + C
sin(y) = e^((2/5)t^(5/2) + C) = Ce^((2/5)t^(5/2))
y(t) = arcsin(Ce^((2/5)t^(5/2))), for y(`pi) = 1
1 = arcsin(Ce^((2/5)`pi^(5/2)))
sin(1) = Ce^((2/5)`pi^(5/2))
C = sin(1)/ e^((2/5)`pi^(5/2)) = approx. .0008
y(t) = arcsin(.0008e^((2/5)t^(5/2)))
@& Good.
Consider the solution for a different initial condition:
tan(y) = sin(y) / cos(y).
Our equation is therefore defined only when cos(y) is not zero, i.e., when y is not equal to pi/2, 3 pi/2 or either of these angles plus a multiple of 2 pi.
Rearranging the equation we get
cos(y) / sin(y) dy = t^(3/2) dt
Integrating we get
ln (sin y)= 2/5 t^(5/2)
Solving for y:
y= arcSin(Ce^(2/5 t^(5/2)))
y(0) = 1 implies
pi / 2 = arcSin( A e^(2.5 * 0^(5/2) ) = arcSin(A)
so that
A = sin(pi/2) = 1.
The solution is therefore
y = arcSin(2/5 t^(5/2) ).
However there is a problem with this solution. y = pi/2 is not in the domain of definition of the function t^(3/2) tan(y).
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`q002. We saw previously that the equation y ' = -.01 t y^2 has solutions of the form y = 1 / (-.005 t^2 - c). For negative values of c this solution has a vertical asymptote at t = sqrt(200 c), and is not defined at t = sqrt(200 c).
We can understand why this equation has vertical asymptotes if we plot the direction field. Plot the direction field defined by t values 0, 2, 4, 6, 8 and 10, and y values 0, 10, 20, 30, 40. Explain what happens when you sketch solution curves from various points of this grid.
The slope approaches neg infinity without bound as you move up and to the right in quad I or larger y and t values
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`q003. The sort of thing that happened with the direction field of y = -.01 t y^2 cannot happen for a linear differential equation of the form y ' + p(t) y = 0, unless p(t) itself is undefined at a point. Explain why this is so.
Because there are not squared terms, as stated only way to be undefined is if there are fractions with variables in denominator which for some value of t or y make the denominator zero which is undefined
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`q004. Write the equation ( y^2 cos(t) ) ' = 0 in the form f(t, y, y') = 0. (To do so, find the derivative of y^2 cos(t) and set it equal to zero).
( y^2 cos(t) ) ' = (y^2)’cos(t) + y^(2)cos(t)’
= cos(t)2y (dy/dt) - y^(2)sin(t)
cos(t)2y (dy/dt) - y^(2)sin(t) = 0
dy/dt = sin(t)y^(2)/cos(t)2y
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Show that this equation is solved by the function y for which y^2 cos(t) = c, where c is an arbitrary constant.
For y^2 cos(t) = c
y^2 = c/cos(t)
y = `sqrt(c/cos(t))
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@& Good, but you need to verify the solution.
The equation becomes
2 y cos(t) y ' - y^2 sin(t) = 0
If y^2 cos(t) = c, then
y = sqrt( c / cos(t)).
y ' = -c sin(t) / cos^2(t) * 1 / (2 sqrt( c / cos(t) ) = c/2 sin(t) / cos^2(t) * sqrt(cos(t) / c) = sqrt(c) / 2 * sin(t) / cos^(3/2)(t)
where the function and its derivative are defined only if cos(t) > 0.
Substituting this function into the equation we get
2 sqrt(c / cos(t)) * sqrt(c) / 2 * sin(t) / cos^(3/2)(t) - c / cos(t) * sin(t) = 0.
Check to see that this simplifies to an identity.
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`q005. Show that the equation
dF = 0,
where F is a function F(t, y), is of the form
N(t,y) dt + M(t, y) dx = 0.
where N(t, y) = F_t and M(t, y) = F_y.
F(t, y) = 2y^2t^2 + 1/y
dF/dt = 4yt^2 + 4ty^2 dy/dt - 1/y^2 dy/dt
dF = 4yt^2 dy + 4ty^2 dt - 1/y^2 dy
For dF = 0 then
4yt^2 dy + 4ty^2 dt - 1/y^2 dy = 0, separating out we get
4ty^2 dt + (4yt^2 - 1/y^2) dy
Of the form N(t,y) dt + M(t, y) dy = 0.
N(t, y) = 4ty^2 dt
M(t, y) = (4yt^2 - 1/y^2) dy
dM/dt = 4t^2 + 2/y^3
dN/dy= 4y^2
M(t, y) = F_y
N(t, y) = F_t
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Show furthermore than N_y = M_t.
N_y = (F_t)_y = F_ty
M_t = (F_y)_t = F_yt
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`q006. Show that the equation
( sqrt(y) - e^t cos(y) + t ) dt + (t / (2 sqrt(y) + e^t sin(y)) dy = 0
is of the form
N dt + M dy = 0, with N_y = M_t.
Of the form N(t,y) dt + M(t, y) dy = 0.
N(t, y) = ( sqrt(y) - e^t cos(y) + t ) dt
M(t, y) = (t / (2sqrt(y) + e^t sin(y)) dy
dM/dt = [2`sqrt(y) + e^t sin(y) - te^t sin(y)]/ [2`sqrt(y) + e^t sin(y)]^2
dN/dy= ½`sqrt(y) - e^t sin(y)
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Integrate N with respect to t, recalling that since y is treated as a constant, the integration constant can be any function f(y).
Int(sqrt(y) - e^t cos(y) + t dt)
= Int(sqrt(y) dt) - Int( e^t cos(y) dt) + Int( t dt)
= (`sqrt(y)t - e^tcos(y) + t^2/2) + f(y)
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Integrate M with respect to y, recalling that since t is treated as a constant, the integration constant can be any function g(t).
Int(t / (2 sqrt(y) + e^t sin(y)) dy)
= t Int( 1/ (2 sqrt(y) + e^t sin(y)) dy)
= t [Int(1/ (2 sqrt(y) dy) + e^t *Int(sin(y)) dy)]
=t[`sqrt(y) + e^t(-cos(y)) + c
= `sqrt(y)t - te^tcos(y) +g(t)
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Show that these integrals can be made equal by making a good choice of f(y) and g(t).
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Let F(t, y) be the common integral of the two functions.
Show that our equation is of the form dF = 0.
`sqrt(y)t - e^tcos(y) ???This is the common intergral correct????
???I’m a little confused about what we are covering I understand what is stated in the book about the reason that the fn n(y) be what is used to calculate derivative wrt t and N(y) is antiderivative of n(y) and it derivative is taken wrt y. The book is kind of of brief about the material but understandable, but when we go over specifics I don’t completely understand everything right off????
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Argue that the solution is therefore of the form F(t, y) = c, for a constant c.
Because
n(y)dy/dt + m(t) = d/dtN(y) + d/dtM(t) = d/dt[N(y) + M(t)] = 0
I know why N(y) + M(t) is a fn of t but not real clear on this part. So N(y) + M(t) involves the unknown solution y = y(t) because N(y) + M(t) were never solved we just know from the chain rule that a connection can be made that the fn’s N(y) + M(t) can be taken with respect to t. I know for whatever reason we cannot be certain of the solution for y = y(t) we get N(y) + M(t) = 0?????????
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@& N dt = ( sqrt(y) - e^t cos(y) + t ) dt
N_y = 1/(2 sqrt y)+ e^t sin y
M dy= (t / (2 sqrt(y) + e^t sin(y)) dy
M_t= 1/(2 sqrt y)+e^t sin y
N_y= M-t= 1/(2 sqrt y)+ e^t sin y= 1/(2 sqrt y)+e^t sin y
Integrating N with respect to t we get
t* sqrt y -e^t cos y +1/2t^2 + f(y).
Integrating M with respect to y we get
t sqrt y - e^t cos y + g(t)
If we let g(t) = 1/2 t^2 and f(y) = 0 the two integrals agree, giving us
F(t, y) = t* sqrt y -e^t cos y +1/2t^2
dF = F_t dt + F_y dy = (sqrt(y) - e^t cos y + t) dt + (t / (2 sqrt(y) + e^t sin(y)) dy
so the equation
dF = 0
is just our original equation
(sqrt(y) - e^t cos y + t) dt + (t / (2 sqrt(y) + e^t sin(y)) dy = 0
The differential of a constant function is 0, and this is the only function whose differential is zero, so
dF = 0
is equilvalent to
F = c.
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&#This looks good. See my notes. Let me know if you have any questions. &#