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course Mth 279
April 3, 2011 around 5pm.
Query 05 Differential Equations*********************************************
Question: 3.2.6. Solve y ' + e^y t = e^y sin(t) with initial condition y(0) = 0.
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Your solution:
Y’ = -e^y* t + e^y*(sin(t))
F_y (partial derivative) = e^(y) * sin((t) -t)
(y’ / -e^(y)*t) - e^(y)*sin(t) = 0
Take the antiderivative:
Int ((y’ / -e^(y)*t) dt) - Int (e^(y)*sin(t) dt) = C
I got stuck here but I know after the integration you solve for y(t) and impose the initial condition y(0) = 0.
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The equation can be written as
dy/dt = e^y ( sin(t) - t )
which we can write as
dy / e^y = (sin(t) - t) dt.
Integrating both sides we get
-e^(-y) = -cos(t) - t^2 / 2 + c,
which we write as
e^(-y) = cos(t) + t^2 / 2 + c
before solving for y. We obtain
y = - ln | cos(t) + t^2 / 2 + c |.
y(0) = - ln | cos(0) + 0^2 / 2 + c | = ln | 1 + c |.
- ln | 1 + c | = 0 if 1 + c = 1, so c = 0 and our solution is
y = - ln | cos(t) + t^2 / 2 |
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Given Solution:
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Self-critique (if necessary):
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Question: 3.2.10. Solve 3 y^2 y ' + 2 t = 1 with initial condition y(0) = -1.
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Your solution:
Y’ = (1-2t) / (3y^2)
F_y = 2(2t-1) / (3y^3)
F_t = (-2) / (3y^2)
Y’*(3y^2) - (1-2t) = 0
Int (y’(t)*(3y^2) dt) - Int (1-2t dt) = C
Int (y’(t)*(3y^2) dt) - (t-t^2) = C
I’m stuck on integrating the left side, but I know how to solve the problem afterwards. Solve for y(t) and impose the initial condition y(0) = -1
@& You need to separate variables. All the y terms on one side, with dy, and all the t terms on the other with dt.
This equation is easily rearranged into the form
3 y^2 dy = (1 - 2 t) dt
which we integrate to get
y^3 = t - t^2 + c.
Solving for y we get
y = (t - t^2 + c) ^ (1/3 ).
Applying the initial condition y(0) = -1 we have
-1 = (0 - 0^2 + c)^(1/3)
so that c = -1.
Our final solution is
y = (t - t^2 - 1) ^ (1/3).
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Given Solution:
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Self-critique (if necessary):
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Question: 3.2.18. State a problem whose implicit solution is given by y^3 + t^2 + sin(y) = 4, including a specific initial condition at t = 2.
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Your solution:
d/dt = 2t
d/dy = cos(y) + 3y^2
Since the implicit solution is found AFTER integration, wouldn’t you find the derivative of the equation, “un-separate” the variables and that will give you the initial value problem?
@& You're certainly on the right track.
This equation can be written as
y^3 + sin(y) = -t^2 + c,
which is the integral of the equation
(3 y^2 + cos(y)) dy = -t dt
so our equation is
(3 y^2 + cos(y) ) y ' = - t.
To get the given implicit solution we need c = 4.
The left-hand side of the equation is equal to 0 when y = 0, and the right-hand side is 0 when t = sqrt(c) (provided c > 0).
Since we need c = 4, our condition becomes y = 0 when t = sqrt(4) = 2, which we express as
y(2) = 0.
ALTERNATIVE SOLUTION
y^3 + t^2 + sin(y) = 4
can be differentiated with respect to t to obtain the differential equation
3 y^2 y ' + 2 t + cos(y) y ' = 0.
y(2) can be determined by letting t = 2 in the equation y^3 + t^2 + sin(y) = 4, obtaining
y^3 + 2^2 + sin(y) = 4 implies
y^3 + sin(y) = 0,
which is satisfied by y = 0. So our initial condition would be
y(2) = 0.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Question: 3.2.24. Solve the equation y ' = (y^2 + 2 y + 1) sin(t) and determine the t interval over which he solution exists.
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Your solution:
Partial derivative: 2(y+1) * sin(t)
How would I finish this using theorem 3.1? (y’ = f(t,y) and y(t0) = y0)
@& Again you need to separate the variable then integrate.
You're trying to find y from information about y and y '. If you take the derivative of the y ' expression you're going in the wrong direction.
We rewrite the equation in the form
dy / (y ^2 + 2 y + 1) = sin(t) dt.
Factoring the denominator of the left-hand side we have
dy / (y + 1) ^ 2 = sin(t) dt
Integrating we get
-1 / (y + 1) = -cos(t) + c
so that
y = 1 / (cos(t) + c) - 1.
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@& See my notes. You should be sure you can work similar problems from the text. I'll be glad to answer questions or provide more explanation.*@