Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 preview

 The required solution is

Step-by-step explanation:   We are given to solve the following differential equation :

Let us consider that

be an auxiliary solution of equation (i).

Then, we have

Substituting these values in equation (i), we get

So, the general solution of the given equation is

Differentiating with respect to t, we get

According to the given conditions, we have

and

From equation (ii), we get

Thus, the required solution is

a) 40 more; b) old factory:   10%, new factory:   16%; c) week 4

explanation:

a) at week 0, we see in the graph that the new factory produces 190 specialty items.  

using the function, substituting 0 in for w,

this is a difference of 230-190 = 40.

b) the function for the old factory is written in the form f(x) = a(1+r)ˣ, where a is the original amount, x is the amount of time and r is the growth rate.   the equation for the old factory is ; 230 is in place of a and is the original amount; w is the amount of time; which means 1+r = 1.1.   this means that r must be 0.1 or 10%.

for the new factory, we find the percent of increase to find the growth rate; the formula for this is

between the first two points, the amount of change is 220-190 = 30; this gives us

30/190 = 0.158 ≈ 0.16 = 16%

c) at week 0, the old factory has 230 while the new factory has 190.

at week 1, the old factory has 230(1.1) = 253 while the new factory has 220.

at week 2, the old factory has 230(1.1²) = 278.3 while the new factory has 252.

at week 3, the old factory has 230(1.1³) = 306.13 while the new factory has 290.

at week 4, the old factory has 230(1.1⁴) = 336.743 while the new factory has 337.   this is the first week that the new factory surpasses the old.

the distance between the school and her home is 3.6km.

12 mins = 0.2 hours = 1/5 hour

54 ÷ 3 = 18km per hour

1/5 of 18 = 3.6km per 12 mins