Question

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.

Answer 1

Answer:

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.