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course Phy 279

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Differential Equations Notes 110209

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Question: `q001. Solve the equation

y ' = sin(t) / y, y(0) = 1.

Over what region(s) of the plane does the existence theorem apply, and what is the domain of your solution?

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Your solution:

y dy = sin(t) dt

y^2/2 = -cos(t) + C

y = `sqrt(-2cos(t) + C)

For y(0) =1

1 = `sqrt(-2 + c)

1 = -2 + c

c = 3

y(t) = `sqrt(-2cos(t) + 3)

Domain

-2cos(t) + 3 must be greater than or equal to 0, -2cos(t) + 3 >= 0

-2cos(t) >= -3

cos(t) <= (3/2) = 1.5, because cos(t) will never be greater than 1 there are no restrictions for the domain

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Question: `q002. In general air resistance has components which depend on v and on v^2, where v is the velocity of an object. If v ' = -5 t + 4 t v + t v^2, what is the solution of the equation for which v(0) = 10?

Over what t interval can we expect this solution to exist?

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Your solution:

v’/(5+4v+v^2) - t = 0

Int(v’/(5+4v+v^2)), letting u= arctan(v + 2), du = 1/(1 + (v+2)^2) dv

dv = (5+4v+v^2) du, plugging back into our equation

Int((1/(5+4v+v^2))* 5+4v+v^2 du) = Int(1 du) = u = arctan(v + 2)

??????I’m not this is the correct approach but I’ve spent some time on this one and this is the best I can come with???????

v’/(5+4v+v^2) - t = 0, after integrals are taken

arctan(v + 2) - t^2/2 = C

arctan(v+2) = t^2/2 + c

v+2 = tan( t^2/2 + c), finally

v = -2 + tan( t^2/2 + c),solution is

10 = -2 + tan(c)

c = arctan(8) = 1.466

v(t) = -2 + tan( t^2/2 + 1.466)

tan fn is not defined at pi/2 + n*pi, where n is any integer

???I don’t believe this is correct, but I can’t see where my error has been?????

@& `

This equation can be arranged in the form dv / (v^2 + 4 v - 5) = t dt. The denominator can be factored.

It is then possible to use partial fractions to integrate the left-hand side.

The integral on the left is 1/6 ln( (v - 1) / (v+5) ), so the solution is given implicitly by

1/6 ln( (v - 1) / (v+5) ) = t^2 / 2 + c

The solution is

v = (5 e^(c + 3 t^2) + 1) / (1 - e^(c + 3 t^2)),

which can be expressed as

v = (5 A e^(3 t^2) + 1) / (1 - Ae^(3 t^2)),

where A > 0 is equal to e^c.

The exponential function is defined for any real value of its exponent, and 3 t^2 is always real when t is real.

This solution is therefore defined as long as the denominator is nonzero, which is the case as long as A e^(3 t^2) is not equal to 1 (you can verify that this is the case as long as 3 t^2 is not equal to ln(1 / A), so that t is not equal to +- sqrt( -1/3 ln(A); it is worth noting that this cannot be the case if A > 1).

If v(0) = 10, then e^(3 t^2) = 1 so

10 = (A + 1) / (1 - A), with solution A = 9/11.

The solution is therefore

v = (45/11 e^(3 t^2) + 1) / (1 - 9/11 e^(3 t^2)) = (45 e^(3 t^2) + 11) / (11 - 9 e^(3 t^2))

As long as the denominator 11 - 9 e^(3 t^2) is not equal to zero, the solution will exist.

11 - 9 e^(3 t^2) = 0 when e^(3 t^2) = 11/9 so that 3 t^2 = ln(11/9) and t = +-1/3 ln(11/9), or approximately t = +- .067.

NOTE STUDENT SOLUTION TO SIMILAR EQUATION

Separated out

v’/(5+4v+v^2) - t = 0

Int(v’/(5+4v+v^2)), letting u= arctan(v + 2), du = 1/(1 + (v+2)^2) dv

dv = (5+4v+v^2) du, plugging back into our equation

Int((1/(5+4v+v^2))* 5+4v+v^2 du) = Int(1 du) = u = arctan(v + 2)

??????I’m not this is the correct approach but I’ve spent some time on this one and this is the best I can come with???????

v’/(5+4v+v^2) - t = 0, after integrals are taken

arctan(v + 2) - t^2/2 = C

arctan(v+2) = t^2/2 + c

v+2 = tan( t^2/2 + c), finally

v = -2 + tan( t^2/2 + c),solution is

10 = -2 + tan(c)

c = arctan(8) = 1.466

???I don’t believe this is correct, but I can’t see where my error has been?????

@& There is a - sign in front of that 5. See the solution appended below.

However, except that it doesn't answer the question, there's nothing wrong with solving the equation as you have it and comparing to the given solution.

arctan(v + 2) is indeed the antiderivative, easily enough obtained by completing the square to put the integrand into the form 1 / (1 + u^2).

For the integration constant you would get the equation

10 = - 2 + tan(c),

but that leads to c = arcTan(12), which would be about 1.52, not arcTan(8), which as you say is about 1.47 .

The solution would therefore be

y = arcTan(v+2) + arcTan(12), approximately

y = arcTan(v+2) + 1.52.

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Question: `q003. Show that the equation

2 y tan(t) dy + (y^2 tan(t) sec (t) + 1 / (2 sqrt(t) ) dt = 0

is of the form M dy + N dt = 0, with M_t = N_y.

Integrate to find the function F(t, y) for which F_t = N, and F_y = M.

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Question: `q004. Consider the equation y ' + p(t) y = q(t) y^3.

Let v(t) = y(t)^m, where the value of m has yet to be determined.

What is dv/dt?

What therefore is dy/dt in terms of v and dv/dt?

What is y in terms of v?

Rewrite the original equation in terms of p(t), q(t), v and dv/dt.

Choose m so that the v in the right-hand side has power 0.

Now the equation is linear nonhomogeneous order 1.

`q What integrating factor allows us to solve the equation?

NOTE: This is an example of a Bernoulli Equation, which is of the form

y ' + p(t) y = q(t) y^n.

This equation differs from a first-order linear homogeneous equation by the factor y^n on the right-hand side.

The general method is to change the variable to v = y^m, then choose the value of m that makes the equation linear.

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Question:

`q005. A population function P(t) satisfies dP/dt = k P, as long as space and resources are unlimited. That is, the rate of growth is proportional to the population.

If, however, the population grows too much, it approaches the carrying capacity L of its environment, and its rate of growth becomes also proportional to (L - P). Thus the rate of population growth is jointly proportional to P and (L - P).

This gives us the equation

dP/dt = k P ( L - P ).

This equation is separable. Find its general solution. You will need to use partial fractions.

If L = 1000, P(0) = 100 and P(1) = 200, what is the value of k?

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@& Check my notes.*@