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course MTH 279
2/10 6:08 pm
110207 differential equationsFrom 110202 questions
`q006. Recall implicit differentiation. If y is a function of t, then what is the derivative with respect to t of the function H(t, y) = t cos y + 1/y?
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H'(t,y)= cos y- t sin y y'+ -1/y^2 y'
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`q007. Verify that the derivative of the function H(t, y) = t^2 sqrt(y) is 2 t sqrt(y) + t^2 / (2 sqrt(y) ) * y '.
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H'(t,y)= d(t^2 sqrt y)/dt = 2t sqrt y + t^2/2 sqrt y * y'
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Verify that the equation
2 t sqrt(y) + t^2 / (2 sqrt(y) ) * y ' = 0
is of the form
( H(t, y) ) ' = 0,
where H(t, y) = t^2 sqrt(y) and ' indicates the derivative with respect to t.
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We know from above that we have H'(t,y)= 2 t sqrt(y) + t^2 / (2 sqrt(y) ) * y ' = 0
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Verify that the equation
( H(t, y) ) ' = 0
is equivalent to the equation
H(t, y) = c,
where c is any constant number.
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Basically all we need to realize here is that dconst/dt= 0.
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For the given function H(t, y) = t^2 sqrt(y), solve the equation
H(t, y) = c
for y as a function of t.
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t^2 sqrt y= c ---> y= (c/t^2)^2= C/t^4
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Show that this function y(t) satisfies the equation
2 t sqrt(y) + t^2 / (2 sqrt(y) ) * y ' = 0.
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Substitution:
2t sqrt (C/t^4)+ t^2/(2 sqrt (C/4^2))*(-4C/t^5)
2tC/t^4 -4Ct^4/2Ct^5= 2C/t-2/t= 0
My constant may have run afoul somewhere but this holds if C= 1
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New questions (separable, exact equations)
`q001. Solve the equation y ' = t^(3/2) tan(y) for the initial condition y(pi) = 1.
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Separate:
cot y dy = t^(3/2) dt
Integrate:
ln (sin y)= 2/5 t^(5/2)
Solve for y:
y= asin(Ce^(2/5 t^(5/2)))
y(π)=1---> C= 7.693x10^-4
I'm not sure that turned out exactly how you intended it to but on the other hand y(1)= π only holds if C= 0.
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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.
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The slopes start of at 0 but grow very steep, negatively, very quickly and they all converge toward the t axis.
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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.
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This is because the solutions to this form of eq is always some exponential function and that is always defined. In other words it's solution doesn't involve terms that are undefined unless p is itself undefined but then the slope field is also undefined and so the behavior is not unexpected like it is for the above equation.
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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).
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2y y' cos t - y^2 sin t =0
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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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Of course it's solved by this function since we just took the derivative of it to get the equation in the first place.
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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) dy = 0.
where N(t, y) = F_t and M(t, y) = F_y.
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N(t,y)= -y^2 sin t
M(t,y)= 2y cos t
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Show furthermore than N_y = M_t.
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N_y= -2y sin t
M_t = -2y sin t
-2y sin t = -2ysin t
@& Good job on the specific example.
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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`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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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
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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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t* sqrt y -e^t cos y +1/2t^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).
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t sqrt y - e^t cos 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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t* sqrt y -e^t cos y +1/2t^2 + f(y) = t sqrt y - e^t cos y + g(t)
Reduces to:
1/2t^2 + f(y) = g(t)
Which means g(t)= 1/2t^2 and f(y)= 0
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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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( sqrt(y) - e^t cos(y) + t ) dt + (t / (2 sqrt(y) + e^t sin(y)) dy = 0
which in notation can be written:
dF= d F/ d t dt + d F/ d y dy = 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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F(t,y)= c
The derivative of c is 0 so this is a solution.
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@& Good job. See my notes.*@