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course Mth 279
10:48 pm 6/23
Query 06 Differential Equations*********************************************
Question: 3.3.4. If the equation is exact, solve the equation (6 t + y^3 ) y ' + 3 t^2 y = 0, with y(2) = 1.
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Your solution:
I'm really unsure about this whole section.
but what I got so far
M = 3t^2 y
N = (6t+y^3)
dM dy = 3t^2
dN dt = 6
the equation is not exact, although I find that hard to believe
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Given Solution:
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Question: If the equation is exact, solve the equation (6 t + 3 t^3 ) y ' + 6 y + 9/2 t^2 y^2 = -t, with y(2) = 1. *
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Your solution:
M = 6y + 9/2 t^2 y^2
N = 6t +3t^3
dM dy= 6 + 2* 9/2 t^2 = 6+9t^2
dN dt = 6 + 9t^2
the equation is exact
then I pick an equality
dH/dy = M = 6 + 9t^2 and integrate with respect to y
dH/dy = 6y + 9t^2 * y
then take the derivative with respect to t
... but that equals zero. ???
#### I think I meant to integrate with respect to t
6t + 9ty^2 = M
then
... integrate by y?
6ty + 3ty^3 = M
then
dH/dy = 6ty + 3ty^3 + dg/dy
then what I just got, plus dg and set equal to N
6ty + 3ty^3 + dg/dy = 6t+ 3t^3
then I think I integrate. But I can't quite figure out which side is with respect to which variable
following the example in the book, it went from
2ty + dg/dy = 2y(t+1)
to
g(y) = y^2 + C_1
But I don't know what they did to get there. ???
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You're on the right track, but you don't quite have the full picture:
We test the equation to see if it is exact, i.e., of the form
M dy + N dt = 0
with M_t = N_x.
The equation can be rearranged to
(t cos(t y) + 2 y e^(y^2)) dy + (y cos(t y) + 1) dt = 0
so it is of the form M dy + N dt = 0 with
M = t cos(t y) + 2y e^(y^2)
and
N = y cos(t y) + 1
M_t = cos(t y) - y t sin(t y)
and
N_y = cos(t y) - y t sin(t y)
These are equal, so our equation is of the form
dF = 0
with
F_y = t cos(t y) + 2y e^(y^2)
and
F_t = y cos(t y) + 1.
F_y = t cos(t y) + 2y e^(y^2) implies that
F = integral( (t cos(t y) + 2 y e^(y^2) ) dy = -sin(t y) + e^(y^2) + g(t),
where g(t) is our integration constant (y being the variable of integration, any function of t has derivative zero with respect to y and so is constant in this integral).
F_t = y cos(t y) + 1 implies that
F = integral( (y cos(t y) + 1) dt) = -sin(t y) + h(y),
where h(y) is constant with respect to t, the variable of integration.
If h(y) = e^(y^2) and g(t) = 0, our result is
F = -sin(t y) + e^(y^2) .
dF = 0 means that
F = c,
where c is constant, so
-sin(t y) + e^(y^2) = c.
The initial condition is that y(0) = pi, so we have
-sin(0) + e^(pi^2) = c
and c = e^(pi^2).
Our implicit solution is therefore given by the equation
-sin(t y) + e^(y^2) = e^(pi^2).
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Given Solution:
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I tried to follow the two examples in the book, but I still don't think I got it
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Question: 3.3.6. If the equation is exact, solve the equation y ' = - ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2) with y(0) = pi.
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Your solution:
M = -(ycos(ty) +1) = -ycos(ty) - 1
N = t cos(ty) + 2ye^(y^2)
uv' + u'v
u = -y, v = cos(ty)
dM dy = -y(-tsin(ty)) + (-1)cos(ty) = ty sin(ty) - cos(ty)
u = t , v = cos(ty)
dN dt = cos(ty) - ty sin(ty) + 0
I think these are supposed to be equal because they are really close. But I tried solving again and still got the same answer.
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The equation is of the form
N(y, t) dy + (t^2 + y^2) sin(t) dt = 0
It is exact if N_t = M_y, where
M_y = 2 y sin(t).
N_t = M_y so that
N = integral(M_y dt) = integral ( 2 y sin(t) dt ) = -2 y cos(t) + g(y),
where g(y) is the most general integration constant for integration with respect to t.
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Given Solution:
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Question: 3.3.10. If N(y, t) * y ' + t^2 + y^2 sin(t) = 0 is exact, what is the most general possible form of N(y, t)?
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Your solution:
N(y,t) y' + y^2 sin(t) = -t^2
integrate
N(y,t) y + 1/3y^3 sin(t) = -1/3 t^3 +C
N(y,t)y = -1/3 t^3 - 1/3 y^3 sin(t) + C
sin(t) acted like a constant because that side was integrated by y
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The equation is of the form
N(y, t) dy + (t^2 + y^2) sin(t) dt = 0
It is exact if N_t = M_y, where
M_y = 2 y sin(t).
N_t = M_y so that
N = integral(M_y dt) = integral ( 2 y sin(t) dt ) = -2 y cos(t) + g(y),
where g(y) is the most general integration constant for integration with respect to t.
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Given Solution:
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Question: 3.3.12. If y = -t - sqrt( 4 - t^2 ) is the solution of the differential equation (y + a t) y ' + (a y + b t) = 0 with initial condition y(0) = y_0, what are the values of a, b and y_0?
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Your solution:
y = -t -sqrt(4- t^2)
y' = -1/2 t^2 + t(4-t^2) ^(-1/2)
plugging in
[(-t -sqrt(4- t^2) ) + at)] (-1/2 t^2 + t(4-t^2) ^(-1/2) ) + ( a (-t -sqrt(4- t^2) ) + bt) = 0
however, this doesn't simplify into anything helpful.
I'm not sure what to do.
I don't need to solve the differential equation because I already have the solution. ???
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Is there an example like this in the book? I know the chapters in mine are different than the way the problems are numbered, but I couldn't find anything
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If y = -t - sqrt( 4 - t^2 ) is the solution of the differential equation (y + a t) y ' + (a y + b t) = 0 with initial condition y(0) = y_0, what are the values of a, b and y_0?
This equation is exact, with M = (y + a t) so that M_y = a, and N = (a y + b t) so that N_t = a.
Thus the equation is of the form dF = 0 with
F = integral( (y + at) dy) = y^2 / 2 + a t y + g(t)
and
F = integral( (a y + b t) dt) = a y t + b t^2 / 2 + h(y).
We reconcile the two forms of F by letting g(t) = b t^2 / 2 and h(y) = y^2 / 2, so that
F = y^2 / 2 + b t^2 / 2 + a t y.
dF = 0 has solution F = c, where c is constant, so the implicit solution is
y^2 / 2 + b t^2 / 2 + a t y = c.
If y = -t - sqrt( 4 - t^2 ) then y^2 = t^2 + (4 - t^2) + 2 t sqrt(4 - t^2) = 4 + 2 t sqr(4 - t^2). Substituting for y in the equation we therefore obtain
4 + 2 t sqr(4 - t^2) + b t^2 / 2 + a t ( -t - sqrt( 4 - t^2) ) = c
(2 t - a t) sqrt( 4 - t^2) + (b/2 - a) t^2 + 4 = c
Thus
2 t - a t = 0
b/2 - a = 0
c = 4
and it follows that
a = 2, b = 4
and when t = 0 we have
y_0^2 / 2 = 4
so that y_0 = 2 sqrt(2).
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The picture you need to understand is this:
An exact equation is an equation of the form d F = 0 for some function F(t, y).
If d F = 0, then F_y dy + F_t dt = 0.
Since (F_y)_t = (F_t)_y, it follows that the t derivative of the expression in front of dt must equal the y derivative of the expression in front of dy.
That is the therefore the test for an exact equation.
Once you have the test straight, it's necessary to differentiate and integrate correctly, and you are still running into problems with those first-year topics. As I've mentioned, practice and review are the only answer to those difficulties.
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