a) Find y as a function of t if y + 12y + 52y = 0, y(0) = 8 , y(0) = 2 . y(t) =

Question

a) Find y as a function of t if y + 12y + 52y = 0, y(0) = 8 , y(0) = 2 . y(t) = ?

b) Find y as a function of t if 25y + 20y + 34y = 0 , y(0) = 9 , y(0) = 6 . y(t) = ?

c) Find y as a function of t if 144y + 312y + 169y = 0 , y(0) = 2 , y(0) = 7 . y(t) = ?

d) Find y as a function of x if y+81y = 0 , y(0) = 2 ,  y(0) = 45,  y(0) = 324 . y(x) = ?

This differential equtions problem calls for the solving of C1 and C2, for all parts. It is from section 4.3 of the differential equtions textbook by Zill. I appreciate any help I can get!!!

Explanation / Answer

y + 12y + 52y = 0, y(0) = 8 , y(0) = 2 . y(t) = ?

let y(t) = e^(rt)
then,
y'(t) = r*e^(rt)
y''(t) = r^2*e^(rt)

Put this in:
y + 12y + 52y = 0
r^2*e^(rt) + 12*(r*e^(rt)) + 52*(e^(rt)) = 0
r^2 + 12r + 52 = 0

r = {-b +/- sqrt (b^2-4ac)} / (2a)
   = {-12 +/- sqrt(144-4*52)}/2
   = {-12 +/- 8i}/2
   = -6+4i   or -6-4i

so,
y(t)= c1*e^(-6t) cos (4t) + c2*e^(-6t) sin (4t)
y(0) = c1*1*1 + c1*1*0
8 = c1

y'(t) = - c1*4*e^(-6t) sin (4t) - c1*6*e^(-6t) cos (4t) + c2*4*e^(-6t) cos (4t) - c2*6*e^(-6t) sin (4t)
y'(0) = -4c1*1*0 - 6*c1*1*1 + 4*c2*1*1 - 6*c2*1*0
2 = -6c1 + 4c2
2 = -6*8 + 4*c2
2 = -48+4c2
c2 = 12.5

y(t)= c1*e^(-6t) cos (4t) + c2*e^(-6t) sin (4t)
y(t) = 8*e^(-6t) cos (4t) + 12.5*e^(-6t) sin (4t)

you can follow the same procedure for others

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