Full credit will be given if you show all analytical work and complete each part
ID: 2969916 • Letter: F
Question
Full credit will be given if you 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: Solve the following IVP: 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: First solve the corresponding homogeneous equation to find one solution. That is, solve Knowing this solution, use reduction of order to obtain the general solution. Check your answer.Explanation / Answer
1)
a)
y'' + 2y' + 8y = 0
its characteristic eq. is
m^2 + 2m + 8 = 0
=>
(m+1)^2 = -7
=>
m = -1 + sqrt(7) i, -1 - sqrt(7) i
solution is
y = e^(-x) * [ A cos(sqrt(7) x) + B sin(sqrt(7) x) ]
put y = sqrt(7), x = 0
sqrt(7) = A
and
y' = - e^(-x) * [ A cos(sqrt(7) x) + B sin(sqrt(7) x) ] + e^(-x) * [-sqrt(7) A sin(sqrt(7) x) + sqrt(7)B cos(sqrt(7) x) ]
put y' = 0, x = 0
0 = - 1 * [ A ] + 1 * [ sqrt(7)B ]
=>
A = sqrt(7) B
=>
sqrt(7) = sqrt(7) B
=>
B = 1
solution is
y = e^(-x) * [ sqrt(7) cos(sqrt(7) x) + sin(sqrt(7) x) ]
b)
x^2y'' + 3xy' + 8y = 0
let z = ln(x) => e^z = x
and we get
xy' = ty
xy'' = t(t-1)y
substitute in the D.E.
t(t-1)y + 3ty + 8y = 0
=>
(t^2 + 2t + 8) y = 0
its characteristic eq. is
m^2 + 2m + 8 = 0 ---[replace t with m]
=>
m = -1 + sqrt(7) i, -1 - sqrt(7) i
solution is
y = e^(-z) * [ C1* cos(sqrt(7) z) + C2* sin(sqrt(7) z) ]
we have e^(z) = x
so, e^(-z) = x^(-1) = 1/x
and we know z = ln(x)
so,
cos(sqrt(7) z) = cos(sqrt(7)*ln(x))
sin(sqrt(7) z) = sin(sqrt(7)*ln(x))
solution is
y = x^(-1) * [ C1* cos(sqrt(7) ln(x)) + C2* sin(sqrt(7) ln(x)) ]
PUT X=1, Y=SQRT(7)
SQRT(7) = [ C1 ] => C1 = SQRT(7)
AND
y' = -x^(-2) * [ C1* cos(sqrt(7) ln(x)) + C2* sin(sqrt(7) ln(x)) ]
+ x^(-1) * [ -SQRT(7)/x *C1* sin(sqrt(7) ln(x)) +sqrt(7) /x * C2* cos(sqrt(7) ln(x)) ]
pu y' = 0, x =1
0 = -1 * [ C1 ]+ [ sqrt(7) * C2 ]
=>
0 = -1 * sqrt(7) + sqrt(7) * C2
=> C2 = 1
solution is
y = x^(-1) * [ sqrt(7)* cos(sqrt(7) ln(x)) + sin(sqrt(7) ln(x)) ]
2)
a)
y'' - 4y = 2e^(2x)
its characteristic eq. is
m^2 - 4 = 0
=>
m = 2, -2
so,
yc = C1*e^(2x) + C2*e^(-2x)
let yp = Axe^(2x)
yp' = 2Axe^(2x) + Ae^(2x)
yp'' = 4Axe^(2x) + 4Ae^(2x)
substitute in the given D.E.
4Axe^(2x) + 4Ae^(2x) - 4Axe^(2x) = 2e^(2x)
compare like terms
4A = 2
=>
A = 1/2
yp = (1/2)*xe^(2x)
solution is
y = yc + yp
= C1*e^(2x) + C2*e^(-2x) + (1/2)*xe^(2x)
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