2. given that y1 (t) = t3 is a solution of the equation 3t2y"-5ty3y = 0 for t > 0 (1) find a second linearly independent solution y2(t) of (1) solution: we proceed by using the reduction of order method. thus, we need to find a function v such that y2(t) v(t)y1(t) is solution of equation (1). if v not a constant function, then y2 is linearly independent (cf lecture notes). then, 2' vuk and inserting this into equation (1) yields 0 3t("y20'y vy{) 5t (v'yn + vy{) -3vyi 3y" + (6y- 5ty1) (3t°y{ - 5tyh -3y)u 3t°y" 13t', where in the third equality, we used that yi is a solution of (1) and that y1(t) = t3. the last equation is equivalent to 13 0 " for t 0 thus w= v is a solution of the 1st order linear ode 13 for t > 0 3 ct13/3 for t 0 and which (by seperation of variables) has the solution w (t) = any constant c e r. thus, v(t) = u cer and so w(s)ds ct10/3 for t > 0 and any constant y2(t) = v(t)y1(t) = t 10/3 f3 since v is a non-constant function, {yı, y2} is linearly independent for t0

i would say c is the best answer, hope it's right!

good luck! =)

cutie~

answer: evaluate the objective function at each vertex to determine which - and -values, if any, maximize or minimize the function. example: find the minimum value and maximum value of the objective function f ( x , y ) = 4 x + 5 y , subject to the following constraints

ou would have 12 of each

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).