In this problem, x = c1 cos t + c2 sin t is a two-parameter family of solutions of the second-order de x'' + x = 0. find a solution of the second-order ivp consisting of this differential equation and the given initial conditions. x(π/3) = 3 2 , x'(π/3) = 0

solution:

we have the solution as

x(t)= c1*cos(t) + c2*sin(t)

=>x'(t) = -c1*sin(t) +c2*cos(t)

given x(π/3)=sqrt(3)/2 and x'(π/3)=0

=>sqrt(3)/2= c1*(1/2) + c2*(sqrt(3)/2) > (1)

and 0= -c1*(sqrt(3)/2) + c2*(1/2) >(2)

=>c2=c1*sqrt(3)

using this in (1) we get

sqrt(3)/2= c1*(1/2) + c1*(3/2)

=>c1= sqrt(3)/4

and c2=3/4

x(t) = (sqrt(3)/4)*cos(t) +(3/4)*sin(t)

i'm sure goog can you get that answer

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.