Query 17

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course MTH 279

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:

y” + 4/5y’ + 4/25y = 0

Assume the solution will take a form of y(t) = e^(λt).

y’(t) = λe^(λt)

y”(t) = λ^2e^(λt)

λ^2e^(λt) + 4/5*λe^(λt) + 4/25*e^(λt) = 0

e^(λt)(λ^2 + 4/5λ + 4/25) = 0

λ^2 + 4/5λ + 4/25 = 0

λ = ((-4/5)±sqrt((4/5)^2-4*(4/25)))/2 = -4/10 = -2/5

Since we have repeated roots, the second complementary solution must be different.

We transform our differential equation as such:

y” - 2(2/5)y’ + (2/5)^2y = 0

Our characteristic polynomial is now:

λ^2 - 2(2/5)λ + (2/5)^2 = (λ-(-2/5))^2

One solution is obviously, as found earlier, y_1 = e^((-2/5)t).

Using the method of reduction of order: y_2 = y_1*u(t) = e^((-2/5)t)*u(t).

y_2’ = -2/5e^((-2/5)t)*u(t) + e^((-2/5)t)*u’(t)

y_2” = 4/25e^((-2/5)t)*u(t)-2/5e^((-2/5)t)u’(t) -2/5e^((-2/5)t)*u’(t) + e^((-2/5)t)*u”(t)

Substitution back into the new DE gives:

y_2” - 2(-2/5)y_2’ + (-2/5)^2y_2 = e^((-2/5)t)*u”(t)

Since the domain of the right hand side is unrestricted, y_2=e^((-2/5)t)*u(t) is a solution only if u”(t) = 0.

ʃd^2/dt^2(u(t))dt = ʃ0dt

u’(t) = A

ʃd/dt(u(t))dt = ʃAdt

u(t) = At+C

Thus, y_2(t) = (At+C)e^((-2/5)t) = Ate^((-2/5)t)+Ce^((-2/5)t) = te^((-2/5)t);

Verifying that y_1 and y_2 form a fundamental set, we compute the Wronskian.

W(t) = |[e^((-2/5)t), te^((-2/5)t);-2/5e^((-2/5)t), e^((-2/5)t)-2/5te^((-2/5)t)]| = e^((-4/5)t)-2/5te^((-4/5)t) + 2/5te^((-4/5)t) = e^((-4/5)t)

The Wronskian is nonzero for all values of t, therefore y_1 and y_2 form a fundamental set. Our general solution is:

y_g = Ae^(-2/5t) + Bte^(-2/5t)

y_g = -2/5Ae^(-2/5t) + B(-2/5te^(-2/5t) + e^(-2/5t))

Imposing our initial conditions:

y(5) = 4e^(-2); y'(5) = -3/5 e^(-2);

y(5) = Ae(-2) + 5Be^(-2) = 4e^(-2)

A = 4 - 5B

y’(5) = -2/5(4 - 5B)e^(-2) + B(-2e^(-2)+e^(-2)) = -3/5e^(-2)

B = 1

A = -1

y(t) = -e^(-2/5t) + te^(-2/5t)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

y” + 2/3sqrt(3)y’ + 1/3y = 0

Without writing through all of the form assumptions, our characteristic equation is:

λ^2 + 2/3sqrt(3)λ + 1/3 = 0

Using the quadratic formula:

λ = (-2/3sqrt(3) ± sqrt((2/3sqrt(3))^2-4/3))/2 = -sqrt(3)/3

We have real repeated roots, so we have one solution of form:

y_1= e^(-t*sqrt(3)/3) and one solution of the form y_2 = e^(-t*sqrt(3)/3)*u(t).

y_2’ = (-e^(-t/sqrt(3))*u)/sqrt(3) + e^(-t/sqrt(3))*u’

y_2” = (e^(-t/sqrt(3))*u)/3 - (2e^(-t/sqrt(3))*u’)/sqrt(3) + e^(-t/sqrt(3))*u”

Substituting y_2, y_2’, and y_2” into the simplified DE and simplifying, we get:

y_2” + 2sqrt(3)/3*y_2’ + 1/3y_2 = e^(-t/sqrt(3))*u”

Since the domain of the exponential portion of the right hand side is unrestricted (exists at all values of t), y_2 = e^(-t/sqrt(3))*u is a solution only if u” = 0.

ʃd/dt(u’)dt = ʃ0dt

u’ = A

ʃd/dt(u)dt = ʃAdt

u(t) = At+C

We can simplify this to say that y_2 = te^(-t/sqrt(3)).

Verifying the solutions make a fundamental set:

W(t) = |[e^(-t/sqrt(3)), t*e^(-t/sqrt(3)); -e^(-t/sqrt(3))/sqrt(3), -t*e^(-t/sqrt(3))/sqrt(3) + e^(-t/sqrt(3))]| = e^(-2t/sqrt(3))

This domain of this Wronskian is unrestricted, and is nonzero at all values of t, so the solutions do form a fundamental set.

y_g = Ae^(-t/sqrt(3)) + B*t*e^(-t/sqrt(3))

y’ = -Ae^(-t/sqrt(3))/sqrt(3) - B*t*e^(-t/sqrt(3))/sqrt(3) + B*e^(-t/sqrt(3))

Imposing our initial conditions:

y(0) = 2 sqrt(3), y ' (0) = 3

y(0) = A = 2sqrt(3)

A = 2sqrt(3)

y’(0) = -(2sqrt(3))/sqrt(3) + B = 3

B = 5

y(t) = 2sqrt(3)e^(-t/sqrt(3)) + 5t*e^(-t/sqrt(3))

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

Since y_1=sin(t), we assume that our second solution takes a form of y_2 = sin(t)u(t).

y_2’ = cos(t)*u + sin(t)*u’

y_2” = -sin(t)*u + cos(t)*u’ + cos(t)*u’ + sin(t)*u” = -sin(t)*u + 2cos(t)*u’ + sin(t)*u”

Plugging this in to our DE, we get:

y_2” - 2cot(t)y_2’ + (1+2cot^2(t))y = sin(t)*u”

Since the domain of the right hand side of the equation is unrestricted, and exists for all values of t, we find that sin(t)*u is a solution if u” = 0.

ʃd/dt(u’)dt = ʃ0dt

u’ = A

ʃd/dt(u)dt = ʃAdt

u = At+C

y_2 = t*sin(t)

We check to make sure that this is a fundamental set:

W(t) = |[sin(t), t*sin(t);cos(t), sin(t) + t*cos(t)]| = sin^2(t)

The Wronskian is nonzero for values of t≠nπ, n = all real integers. Since it is zero for some values of t, we have to dive a little bit deeper.

c(sin(t)) + k(t*sin(t)) = 0

Since there is no non-zero constant combination of c and k that yield a value of 0, we know that these solutions are linearly independent, and form a fundamental set.

y_g = Asin(t) + Bt*sin(t)

This solution exists everywhere, and is continuous at all real values of t.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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Self-critique (if necessary): OK

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&#Good work. Let me know if you have questions. &#