Question

The solution of a certain differential equation is of the form y(t)=a exp(3t) + bexp(8t), where a and b are constants.
The solution has initial coniditons y(0) and y’(0)=1.
Find the solution by using the initial conditions to get linear equations fro a and b
y(t)=?

Answer 1

plz mention the value of y(0)

Answer 2

Since y(0) = 1,

1 = a + b

Since y'(0) = 1,

1 = 3a + 8b

Solve these simultaneously!

From the first equation, a = 1 - b, so

1 = 3(1 - b) + 8b

1 = 3 - 3b + 8b

-2 = 5b, b = -2/5

Therefore, a = 1 - b = 1 + 2/5 = 7/5

Answer 3

given y(t) = aexp(3t) + bexp(8t)

differentiating y(t) with respect to t , we have,

dy/dt =3aexp(3t) + 8bexp(8t)

given y(0) = 1 which means y(t) = 1 when t=0

so substituting t=0 in y(t) , we get

a + b = 1 ---------<1>

also given y’(0) = 1 , so when t=0 in dy/dt we have

3a +8b = 1 ------<2>

solving <1> and <2>, we have b = (-2/5) and a = 1 - b = 7/5 ,

which when substituted in given y(t) gives the required solution as

y(t) = (7/5)exp(3t) + (-2/5)exp(8t).