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course Phy 279
3/6 3Had some trouble with the Exact differential Equations any comments would be greatly appreciated.
Query 06 Differential Equations
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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:
M(t, y) = 3t^2y
N(t, y) = 6 t + y^3
`dM/`dy = 3t^2
`dN/`dt = 6
So this equation is not exact
@& Good.
For reference:
The equation
(6 t + y^3 ) y ' + 3 t^2 y = 0
can be written as
(6 t + y^3 ) dy + 3 t^2 y dy = 0,
and is of the form M dy + N dt = 0 with
M = 6 t + y^3
N = 3 t^2 y.
If there exists a function F such than dF = M dy + N dx, then the equation is of the form dF = 0, with solution F = constant. dF = F_y dy + F_t dt, so this is the case if there exists a function F such that F_y = M and F_x = N.
In this case case if M_t = F_yt = F_ty = N_y, so the condition is
M_t = N_y.
However for our current functions M and N we have
M_t = 6
and
N_y = 6 y t.
The two are not equal, so there exists no such function F.
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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(t, y) = 6 y + 9/2 t^2 y^2
N(t, y) = 6 t + 3 t^3
`dM/`dy = 6 + 9t^2y
`dN/`dt = 6 + 9t^2
????I’m doing something wrong because all my equations are coming out to be not exact and I’m sure that some of them are supposed to be?????????
So this equation is not exact
@& Not necessarily, and not so in the case of these two equations.
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@& Also for reference:
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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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:
y ' = ( y cos(t y) + 1) / (t cos(t y) + 2 y e^y^2)
(t cos(t y) + 2 y e^y^2)y ' - y cos(t y) - 1 = 0
M(t, y) = - y cos(t y) - 1
N(t, y) = t cos(t y) + 2 y e^y^2
`dM/`dy = cos(t y)
`dN/`dt = cos(t y)
H(t, y): dH/dt = M(t, y)= -ycos(ty) -1 and dH/dy = N(t, y)= t cos(t y) + 2 y e^y^2
antidifferenting dH/dy = t cos(t y) + 2 y e^y^2
H(t, y)= sin(ty) + e^(y^2) + g(t)
To determine g(t) , take dH/dt = d/dt[sin(ty) + e^(y^2) + g(t)]
d/dt[sin(ty) + e^(y^2) + g(t)] = ycos(ty) + dg/dt
Following conditions dg/dt = -1 + C_1
So that H(t, y)
H(t, y)= sin(ty) + e^(y^2) - 1
sin(ty) + e^(y^2) - 1 = C, imposing init cond
C = e^(`pi^2) - 1
sin(ty) + e^(y^2) - 1 = e^(y^2) - 1
sin(ty) + 1 = 1
????I believe I messed up because sin^-1(0) = 0, and 0 * t = 0 and y shouldn’t equal 0???????
@&
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: 3.3.10. 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(t, y) = N(t, y)
M(t, y) = t^2 + y^2sin(t)
dH/dt = t^2 + y^2 sin(t)
H(t, y) = (1/3)t^3 + y^2cos(t) + g(y)
dH/dy = 2ycos(t) + dg/dt
2ycos(t) + dg/dt = N(t, y)
N(t, y) = 2ycos(t) + dg/dt
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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 + a t) y ' + (a y + b t) = 0
M(t, y) = (a y + b t)
N(t, y) = (y + a t)
`dM/`dy = ay
`dN/`dt = at
H(t, y): dH/dt = M(t, y)= (a y + b t)and dH/dy = N(t, y)= (y + a t)
antidifferenting dH/dy = (y + a t)
H(t, y)= (y^2/2 + a ty) + g(t)
To determine g(t) , take dH/dt = d/dt[(y^2/2 + a ty) + g(t)]
d/dt[sin(ty) + e^(y^2) + g(t)] = ay + dg/dt
Following conditions dg/dt = C_1
So that H(t, y)
H(t, y)= y_0^2/2 + a*0*y = 1, y_0 = `sqrt(2)
solving for a we get a = 2
???I made a mistake somewhere ??
@&
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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@& Check out my notes and let me know if you have questions, or if you disagree with anything.*@