Differential Equations: COMPLEX ROOTS (((( Two questions am I completing this co

Question

Differential Equations: COMPLEX ROOTS

(((( Two questions am I completing this correctly up to this point? And can someone complete so I have something to check my answer against....)))))

Find y as a function of t if

16y"-24y'+34y=0, y(3)=2, y'(3)=2

y= ANSWER.........

I started this problem and it quickly got out of control.

So what I got so far is...

16r^2 -24r + 34 = 0

[24 +/- sqrt((-24)^2-4(16)(34))]/(2(16)) <------ quadratic equation

(3/4)+/-[sqrt(576-2176)]/32

r=3/4, +/-(5/4) making lambda=(3/4) and mu=(5/4)

writting down the solution format and its derivative

y=exp((3/4)t)[C1*cos((5/4)t)+C2sin((5/4)t)]

y'=exp((3/4)t)[(-5/4)C1*sin((5/4)t) + (5/4)C2*cos((5/4)t)] +

(3/4)exp((3/4)t)[C1*cos((5/4)t)+C2*sin(5/4)t)]

at this point i was alittle shaky about my solution so far... Then the hammer dropped...

putting y(3)=2 into the equation

2=exp(9/4)[C1*cos(15/4)+C2*sin(15/4)]

-----Normally this is where there would be a zero for t and monsters dissapear.

So solving for C1 I got

C1 = 2/[exp(9/5)*cos(15/4)] + [(C2*sin(15/4))/(cos(15/4))]

Before I go any further, because substituting for C1 in the y'(3)=2 equation seems like a retarded amount of work for a question worth 1 pt. my two questions are,

"Am I doing this correctly?" and "Can someone please solve it so I can check my answer with theirs?"

Explanation / Answer

we have

16r^2 - 24r + 34 = 0

r = [24 + sqrt(576 - 2176)]/32, [24 - sqrt(576 - 2176)]/32

= [24 + sqrt(-1600)]/32, [24 - sqrt(-1600)]/32

= [24 + 40 i]/32, [24 - 40 i]/32

= 3/4 + i 5/4, 3/4 - i 5/4

solution is

y = e^(3t/4) * [ A cos(5t/4) + B sin(5t/4) ]

put y = 2, t = 3

2 = e^(9/4) * [ A cos(15/4) + B sin(15/4) ]    -----------eq.1

y' = (3/4) *e^(3t/4) * [ A cos(5t/4) + B sin(5t/4) ] + e^(3t/4) * [ -(5A/4) sin(5t/4) + (5B/4) cos(5t/4) ]

put y' = 2, t = 3

2 = (3/4) *e^(9/4) * [ A cos(15/4) + B sin(15/4) ] + e^(9/4) * [ -(5A/4) sin(15/4) + (5B/4) cos(15/4) ]

2 = e^(9/4) * [ (3A/4) cos(15/4) + (3B/4) sin(15/4) -(5A/4) sin(15/4) + (5B/4) cos(15/4) ]

2 = e^(9/4) * [ cos(15/4) * ((3A/4) + (5B/4) ) + sin(15/4) * ( (3B/4) - (5A/4) ) ] ----eq.2

we get

B = e^(-9/4) * [2 + 10tan(15/4)] / [5cos(15/4) + 5tan(15/4)sin(15/4)]

= e^(-9/4) * [2cos(15/4) + 10sin(15/4)] / 5

and

A = 2e^(-9/4)*sec(15/4) - e^(-9/4) *[2sin(15/4) + 10sin(15/4)tan(15/4)] / 5

solution is

y = e^(3t/4) * [ (2e^(-9/4)*sec(15/4) - e^(-9/4) *[2sin(15/4) + 10sin(15/4)tan(15/4)] / 5) *cos(5t/4)

+ (e^(-9/4) * [2cos(15/4) + 10sin(15/4)] / 5 ) sin(5t/4) ]

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.