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course MTH 279
12:35 pm 2/28
Query 06 Differential Equations*********************************************
Question: 3.3.4. Solve the equation (6 t + y^3 ) y ' + 3 t^2 y = 0, with y(2) = 1.
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Your solution:
This eq is not exact, and not separable, since:
M = 3t^2 y
N =6t + 3y
M_y = 3t^2
N_t = 6
M_y≠ N_t
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Question: 3.3.6. 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:
In form of Exact:
(t cos ty + 2y e^y^2) y' + (y cos ty + 1) = 0
M_y = cos ty - y^2 sin ty
N_t = cos ty - t^2 sin ty
Again these are not equal so the eq is not exact or separable.
@& 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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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 = M_y = 2y sin t
F_y = 2 y sin t --> F= y^2 sin t + h(t)
F_t = t^2 + y^2 sin t --> F = 1/3t^3 + y^2 sin t + g(y)
Thus generally:
N = 2y sin t
@& Good, but you can have a function g(y) as integration constant:
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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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:
F_y = y+ at --> F = 1/2y^2 + yat + g(t)
F_t = ay + bt --> F = aty + 1/2bt^2 + h(y)
g(t) = 1/bt^2
h(y) = 1/2y^2
1/2y^2 + aty + b/2t^2 = C
Here y should be able to solve for y to obtain the given solution but I can't seem to do that but I think the constants can be found anyway.
a= -1, b = 2, y_0 = -2
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@& Check the following and see if you agree.
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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