Question

Consider the differential equation y'' − y' − 12y = 0. Verify that the functions e−3x and e4x form a fundamental set of solutions of the differential equation on the interval (−[infinity], [infinity]). The functions satisfy the differential equation and are linearly independent since the Wronskian W e−3x, e4x = $$7ex ≠ 0 for −[infinity] < x < [infinity].

Answer 1

Answer:

e^(-3x) and e^(4x) form a fundamental set of solutions of the differential equation

y'' - y' - 12y = 0.

Since their wronskian is 7e^x and not 0

Step-by-step explanation:

Given the differential equation

y'' - y' - 12y = 0.

We are required to verify that the functions e^(-3x) and e^(4x) form a fundamental set of solutions of the differential equation on the interval (−[infinity], [infinity]). They form a fundamental set if they are linearly independent, and they are linearly independent if their wronkian is not zero, otherwise, they are linearly dependent.

Now, we need to find the Wronskian of e^(-3x) and e^(4x), and see if it is equal to zero or not.

The wrinkles of two functions y1 and y2 is given as the determinant

W(y1, y2) = |y1...........y2|

......................|y1'.........y2'|

W(e^(-3x), e^(4x))

= |e^(-3x)............e^(4x)|

...|-3e^(-3x)......4e^(4x)|

= e^(-3x) × 4e^(4x) - (-3e^(-3x) × e^(4x))

= 4e^x + 3e^x

= 7e^x

≠ 0

Since the wronskian is not zero, we conclude that the are linearly independent, and hence, form a set of fundamental solution.