Please do qeustion #2. Please follow the guidelines given in the question. That

Question

Please do qeustion #2.

Please follow the guidelines given in the question.

That is, all steps for reduction of order.

I also need part (c) .

Will not take as an answer if your answer fails to meet the above.

Thanks,

Full credit will be given if yon show all analytical work and complete each part of the question as well as turn in your code and the corresponding figures. The correct figure is given in Figure 1 on the next page. Solve the following IVP: y" + 2y + 8y = 0, y (0) = 7y'(0) = 0. Solve the following IVP: x2y" + 3xy' + 8y = 0, y (1) = 7,y'(1) = 0. Plot the solutions from parts 1 and 2 on the same figure. Plot the solution from the first equation on the interval [0,4pi] and the solution from the second equation on the interval [1,4pi]. Comment on the behavior of the different solutions to each IVP. How do the solutions differ in behavior? Why is this? Solve the following ODE: y"-4y = 2e2x. First solve the corresponding homogeneous equation to find one solution. That is, solve y" - 4y = 0. Knowing this solution, use reduction of order to obtain the general solution. Check your answer.

Explanation / Answer

a) k^2 - 4 = 0

k = 2 or -2

y = c1*e^(2x) + c2*e^(-2x)

b) We assume y_p = Axe^(2x)

y" - 4y = 2e^(2x) = A(4e^(2x))

A = 1/2

y = c1e^(2x) + c2e^(-2x) + (x/2)*e^(2x)

c) y' = 2c1e^(2x)-2c2e^(-2x) + xe^(2x) + 0.5*e^(2x)

y" = 4c1e^(2x) -4c2e^(-2x) + 2xe^(2x) + e^(2x) + e^(2x)

y" - y' = 4c1e^(2x) -4c2e^(-2x) + 2xe^(2x) + e^(2x) + e^(2x) - 4(c1e^(2x) + c2e^(-2x) + (x/2)*e^(2x))

= 4c1e^(2x) -4c2e^(-2x) + 2xe^(2x) + e^(2x) + e^(2x) - 4c1e^(2x) + 4c2e^(-2x) + (2x)*e^(2x)

= 2*e^(2x)

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