Question

Find two linearly independent solutions of 22′′−f′(−1,1)=0, f′′(−1,1)>0 of the form:

a) 1=1(1,1,2,2,3,… ) 2=2(1,1,2,2,3,… )
b) 1=2(1,1,2,2,3,… ) 2=1(1,1,2,2,3,… )
c) 1=1(2,2,4,4,6,… ) 2=2(2,2,4,4,6,… )
d) 1=1(1,1,2,2,3,… ) 2=1(2,2,4,4,6,… )

Answer 1

The two linearly independent solutions of 22′′−f′(−1,1)=0, f′′(−1,1)>0 of the form: a) 1 = 1(1,1,2,2,3,… ) 2 = 2(1,1,2,2,3,…).

Step-by-step explanation:

To find two linearly independent solutions of the given differential equation, we need to find two different functions that satisfy the equation.

Let's start by considering the general form of the solution:

f(t) = Aekt

where A and k are constants.

We are given that f''(-1,1) = 0 and f''(-1,1) > 0.

To satisfy these conditions, we can choose two different values for k:

1) k = 1

In this case, the first solution is f(t) = Aet.

The second solution can be found by considering a different choice of constants:

2) k = 2

Here, the second solution is f(t) = Be2t.

Therefore, the correct choice is a) 1 = 1(1,1,2,2,3,… ) 2 = 2(1,1,2,2,3,…). These solutions are linearly independent because they have different exponents in the base e.