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Mth 279
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Brief Question on Query 06
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In Query 06, this was a question I submitted and you responded to.
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
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M_y for this equation is 6 + 9 t y, whereas N_x is 6 + 9 t. The two aren't quite equal so the equation is not exact.
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Let's change the equation to make it exact. The new equation will be
(6 t + 3 t^3 ) y ' + 6 y + 9 t^2 y^2 = -t
For this equation, if the form is
N dy + M dx = 0
we have
N = 6 t + 3 t^3
and
M = 6 y + 9 t^2 y + t
You can check now that N_t = M_y and the equation is exact (both derivatives will be equal to 6 + 9 t^2).
So the equation is of the form
dF = 0
i.e.,
F_y dy + F_x dx = 0
with
F_y = N and F_t = M.
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To get F, then, we integrate N with respect to y and M with respect to t, and reconcile our integrals.
The integral of N = 6 t + 3 t^3 with respect to y is
(6 t + 3 t^3 ) y + g(t),
where g(t) is the integration constant. (The integration constant can be a function of t since the derivative of such a function with respect to y is zero).
The integral of M = 6 y + 9 t^2 y + t with respect to t is
6 t y + 3 t^3 y + t^2 / 2 + h(y),
where h(y) is constant with respect to t.
Thus
F(t, y) = (6 t + 3 t^3 ) y + g(t) = 6 t y + 3 t^3 y + t^2 / 2 + h(y)
so that
6 t y + 3 t^3 y + g(t) = 6 t y + 3 t^3 y + t^2 / 2 + h(y).
Two terms on both sides match, so the equation simplifies to
g(t) = t^2 / 2 + h(y).
This works fine if g(t) = t^2 / 2 and h(y) = 0.
Our F(t,y) function is therefore
F(t, y) = 6 t y + 3 t^3 y + t^2 / 2.
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Now our premise was that the equation is of the form
dF = 0.
The differenial of a constant function is zero, so we conclude that
F = c
for some constant C
This leads to the equation
6 t y + 3 t^3 y + t^2 / 2 = c.
We don't expect that our equation will always be solvable for y in terms of t, and when it isn't our equation defines the solution y(t) implicitly. However for this function we can solve for y. We obtain
y = (t^2 / 2 + c) / (6 t + 3 t^3).
c is our arbitrary integration constant, which we could adjust to satisfy a given initial condition. Since no initial condition is given, though our solution stands as it is.
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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.
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I've removed the rest of the document, because I don't know where those sines and cosines came from either. I was probably looking at a different problem when I wrote that.
See my notes above.
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