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course Mth 279
3/ 25 16
Query 17 Differential Equations*********************************************
Question: Solve the equation
25 y '' + 20 y ' + 4 y = 0, y(5) = 4 e^-2, y ' (5) = -3/5 e^-2
with y(5) = 4 e^-2 and y ' (5) = -3/5 e^-2.
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Your solution:
We can set up a characteristic equation of
25 r^2 + 20 r + 4 = 0
This can be simplified to
(5 r + 2)^2 = 0
r = - 2 / 5
The general solution is
y(t) = c1 e^(-2/5 t) + c2 e^(-2/5 t)
With initial conditions y(5) = 4 e^-2 and y(5) = -3 / 4 e^ - 2
y(5) = c1 e^-2 + c2 e^-2 = 4 e^-2
y(5) = -2 / 5 c1 e^-2 - 2 / 5 c2 e^-2
dividing the y(5) equation by e^-2 we have
c1 + c2 = 4
dividing the y(5) equation by e^-2 and then multiplying both sides by 5 we have
-2 c1 - 2 c2 = -3
???? I cannot find a value to satisfy both equations, even when using woflram alpha
.
I do not see where I went wrong above.
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You have a repeated root r = -2/5, giving you one solution e^(-2t/5 ) and another linearly independent solution. You used e^(-2/5 t) as your second solution, .but in fact c1 e^(-2/5 t) + c2 e^(-2.5 t) is just a single function (c1 + c2) e^(-2.5 t). (c1 + c2) is just a single arbitrary constant, since c1 and c2 are arbitrary in the first place.
Your second linearly independent solution would be t e^(-2t / 5).
Using these two solutions to the general equation, you will be able to satisfy the two initial conditions.
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Question: Solve the equation
3 y '' + 2 sqrt(3) y ' + y = 0, y(0) = 2 sqrt(3), y ' (0) = 3
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Your solution:
The characteristic equation is
3 r^2 + 2 sqrt(3) r + 1 = 0
Using the quadratic equation r is a repeated solution
r = -sqrt(3) / 3
The fundamental set is
[ e^(-sqrt(3) / 3 t) , t e^(-sqrt(3) / 3 t)]
The general solution is
y(t) = c1 e^(-sqrt(3) / 3 t) + c2 t e^(-sqrt(3) / 3 t)
With initial conditions
y(0) = c1 = 2 sqrt(3)
y(0) = - sqrt(3) / 3 c1 + c2 = 3
-6 + c2 = 3
c2 = 9
The solution is
y(t) = 2 sqrt(3) e^(-sqrt(3) / 3 t) + 9 t e^(-sqrt(3) / 3 t)
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The derivative of this expression, evaluated at t = 0, is 2 + 9 = 11, not 3.
c1 = 2 sqrt(3) so y ' (0) = 2 sqrt(3) * -sqrt(3) / 3 = -2. So you get
-2 + c1 = 3
giving you c1 = 5.
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Question: Solve the equation
y '' - 2 cot(t) y ' + (1 + 2 cot^2 t) y = 0,
which has known solution y_1(t) = sin(t)
You will use reduction of order, find intervals of definition and interval(s) where the Wronskian is continuous and nonzero. See your text for a more complete statement of this problem.
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Your solution:
y2(t) = y1(t) u(t)
y2(t) = sin(t) u(t)
(sin(t) u(t)) - 2 cot (t) (sin(t) u(t)) + (1 + cot^2 t) (sin(t) u(t)) = 0
y - 2 cot(t) y + (1 + 2 cot^2 t) y = sin(t) u(t)
u = 0
u(t) = A1 t + A2
y2(t) = sin(t)(A1 t + A2)
y2(t) = A1 t sin(t) + A2 sin(t)
If we set A2 = 0 then
y2(t) = A1 t sin(t)
y(t) = A sin(t) + B t sin(t)
Computing the Wronksian to determine if y1 and y2 are a fundamental set
W(t) = [f(t) g(t)]
f(t) = sin(t)
f(t) = cos(t)
g(t) = t sin(t)
g(t) = sin(t) + t cos(t)
W(t) = f(t)g(t) - g(t)f(t)
W(t) = sin^2 (t) + t sin(t) cos (t) - t sin(t) cos(t) = sin^2 (t)
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It doesn't appear that you used the product rule to find the derivatives of y_2 = sin(t) u(t).
We begin by letting y_2(t) = u(t) * y_1(t), so that
y_2 ' = u ' * y_1 + u * y_1 ' = u ' sin(t) + u cos(t)
and
y_2 '' = u'' * y_1 + 2 u ' * y_1 ' + u y_1 ''
= u '' * sin(t) + 2 u ' * cos(t) - u * sin(t).
Our equation becomes
u'' * y_1 + 2 u ' * y_1 ' + u y_1 '' - 2 cot(t) (u ' * y_1 + u * y_1 ') + (1 + 2 cot^2(t)) * u y_1 = 0
which can be rearranged to yield
[ u y_1 '' - 2 cot(t) * u y_1 ' + (1 + 2 cot^2(t)) * u y_1 ] + u '' y_1 + u ' ( 2 y_1 - 2 cot(t) y_1) = 0.
The terms in brackets can be expressed as u ( y_1 '' - 2 cot(t) * y_1 ' +(1 + 2 cot(t)^2) y_1); since y_1 is a solution to our original equation these terms therefore add up to zero.
This leaves us with
u '' + 2 u ' y_1 (1 - cot(t)) = 0.
or substituting y_1 = sin(t)
u '' + 2 u ' sin(t) ( 1 - cot(t)) = 0.
cot(t) = cos(t) / sin(t), and our equation for u becomes
u '' + 2 (sin(t) - cos(t) ) u' = 0.
Letting v = u ' our equation is
v ' + 2 ( sin(t) - cos(t) ) v = 0.
This is a first-order linear equation, solvable for v. The solution is integrated to find u, and our solution is y = u * y_1 = u * sin(t).
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&#Good work. See my notes and let me know if you have questions. &#