Consider the initial value problem 25y′′+40y′+16y=0, y (0) = a, y′ (0) = -1. Find the critical value of a that separates solutions that become negative from those that are always positive for t>0.

Step-by-step explanation:

Given the differential equation

25y′′+40y′+16y=0

Using D operator to find the complementary solution

Since the Differential equation is equal to 0, then it doesn't have a partial solution.

25y′′+40y′+16y=0

25D² + 40D + 16 = 0

25D² + 20D + 20D + 16 = 0

5D (5D+4) + 4 (5D+4) = 0

(5D+4) = 0 twice

D = - 4/5 twice,

So the solution is a real and equal roots

y (t) = (A+B•t) exp (-0.8t)

Where A and B are constant

The initial value are.

y (0) = a, y′ (0) = -1

y (0) = a = (A+B (0)) exp (-0.8 (0))

a = A

Then, the constant A = a.

Now, y' (t)

y' (t) = B•exp (-0.8t) - 0.8 (A+B•t) exp (-0.8t

-1 = B - 0.8A

Since A = a

Then, B = 0.8a - 1

The solution becomes

y (t) = (a + (0.8a-1) •t) exp (-0.8t)

For t>0

For positive,

a + (0.8a - 1) > 0

1.8a > 1

a > 1/1.8

a > 10/18 > 5/9

a > 5/9.