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course Phy 279
3/6 3I have not submitted this assignment but have covered the material in other assignments???
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Differential Equations Notes 110207
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Question:
`q001. Solve the equation y ' = t^(3/2) tan(y) for the initial condition y(0) = pi/2.
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Your solution:
dy/dt = t^(3/2) tan(y)
dy/tan(y) = t^(3/2) dt, taking integrals of both sides
Int( 1/tan(y) dy) = Int((cosy/siny) dy), u = siny, du = cosy
= Int((1/u) dy) = ln|u| = ln|sin y|
ln|sin y| = (2/5)t^(5/2) + C
sin y = e^((2/5)t^*(5/2) + C) = Ae^((2/5)t^*(5/2))
For y(0) = pi/2
1 = Ae^0 = A = 1
sin y = e^((2/5)t^*(5/2))
???We wouldn’t need to take the arcsin of e^((2/5)t^*(5/2) right????
@& When possible you should get an explicit solution. Your solution is implicit, but not explicit.
In other words you should do the arcSine.:
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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Question: `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.
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Your solution:
We get larger and larger neg values for the slope. So that there appears to be a vertical asymptote approached.
????Is that the same as saying the slope approaches neg infinity???????
@& `
Whether +infinity or -infinity can depend on the direction of your approach, which depends on your initial point.*@
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Question: `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.
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Your solution:
Because our general solution is y = Ce^(-P(t)) , which is defined as long as P(t) has an antiderivative. Also we don’t have fractions to worry about when solving for y unless they are in the exponent of e.
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Question:
`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).
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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.
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Your solution:
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Given Solution:
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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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Question: `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.
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Show furthermore than N_y = M_t.
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Your solution:
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To show that this is the case for any function F:
If N = F_t and M = F_y, then since
dF = F_t dt + F_y dy
our equation is of the given form.
In that case,
N_y = (F_t) y = F_ty, and
M_t = (F_y)_t = F_y t.
Since the order of partial derivatives doesn't matter we have
F_ty = F_yt,
so that N_y = M_t.
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Question:
`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.
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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).
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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).
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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.
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Argue that the solution is therefore of the form F(t, y) = c, for a constant c.
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Your solution:
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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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Given Solution:
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@& You're on the right track.
Check my notes.*@