#$&*
course Mth 279
April 30 around 4:10pm.
Query 16 Differential Equations*********************************************
Question: Find the general solution to
y '' - 5 y ' + 2 y = 0
and the unique solution for the initial conditions y (0) = -1, y ' (0) = -5.
How does the solution behave as t -> infinity, and as t -> -infinity>?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
The general solution of fundamental sets of solutions: y(t) = c1y1(t) + c2y2(t)
Since y + p(t)y + q(t)y = 0 and y(t0) = y0 and y(t0) = y0
and
y(t0) = c1y1(t0) + c2y2(t0) = y0
y(t0) = c1y1(t0) + c2y2(t0) = y0
Our equation: y '' - 5 y ' + 2 y = 0
p(t) = (-5)
q(t) = 2
y(0) = c1y1(0) + c2y2(0) = -1
y(0) = c1y1(0) + c2y2(0) = -5
I need to find y1(t) and y2(t), how?
After I find these, I just plug into equations and see what the system reduces to.
See what c1 and c2 is and then see what the solution of the initial value problem y(t) is and the interval.
@& You find y_1 and y_2 by first solving the characteristic equation.
The characteristic equation is
r^2 - 5r + 2 =0
with solutions
r = 1/2(5 ± sqrt 17)
yielding solution set { e^((5 + sqrt(17) ) / 2 * t), e^((5 - sqrt(17) ) / 2 * t) }/
The general solution can be expressed as
y = c1*e^(1/2(5+ sqrt 17)t) + c2*e^(1/2(5- sqrt 17)t)
y(0)= c1+ c2 = -1 and
y'(0)= c1*(1/2(5+ sqrt 17) + c2 *(1/2(5- sqrt 17) = -5.
Solving these two equations simultaneously, the exact values of c1 and c2 are
c1 = -1/2 - 5 sqrt(17) /34
c2 = -1/2 + 5 sqrt(17) /34
Approximate values are
c1 = -1.1
c2 = 0.1
yielding approximate solution
y = -1.1e^(1/2(5+ sqrt 17)t) + 0.1 *e^(1/2(5 - sqrt 17)t)
As t -> infinity the exponents of both both terms remain positive and become large, with the negative term having the greater magnitude; as a result the limit is -infinity.
As t -> - infinity both exponents approach -infinity so both terms approach zero, resulting in a limit of zero.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Question: Find the general solution to
8 y '' - 6 y ' + y = 0
and the unique solution for the initial conditions y (1) = 4, y ' (1) = 3/2.
How does the solution behave as t -> infinity, and as t -> -infinity>?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
I would solve this one exactly like I solved the first one. If I did everything right
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@&
Our
characteristic equation is
8 r^2 - 6 r + 1 = 0
with solutions
r = (6 +- sqrt(6^2 - 4 * 8 * 1) ) / (2 * 8)
= (6 +- 2) / 16 = 1/2 or 1/4.
Thus our fundamental set is
{e^(1/2 t), e^(1/4 t) }
and out general solution is
y = c_1 e^(t/2) + c_2 e^(t / 4)
y(1) = 4 and y ' (1) = 3/2 yield the equations
c_1 e^(1/2) + c_2 e^(1/4) = 4
1/2 c_2 e^(1/2) + 1/4 c_2 e^(1/4) = 3/2
We solve for c_1 and c_2 to obtain our solution.
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