Final answer:
To find the functions that satisfy the differential equation y''-y'-30y=0 and are linearly independent, we can use the method of finding the characteristic equation. The characteristic equation is obtained by substituting y = e^rx into the differential equation. By solving the characteristic equation and plugging in the values of r into y = Ae^rx and y = Be^sx, we can find the linearly independent solutions to the differential equation. However, in this case, the Wronskian of e^-5x and e^6x is found to be 0, indicating that the two functions are linearly dependent and not linearly independent.
Step-by-step explanation:
The given differential equation is y ′′−y ′ −30y=0. To find the functions that satisfy this equation and are linearly independent, we can use the method of finding the characteristic equation. The characteristic equation is obtained by substituting y = erx into the differential equation. Solving the characteristic equation will give us the values of r. By plugging in the values of r into y = Aerx and y = Besx, we can find the linearly independent solutions to the differential equation.
In this case, the characteristic equation is r2 - r - 30 = 0. By factoring or using the quadratic formula, we can find the values of r to be r = -5 and r = 6. Therefore, the linearly independent solutions to the differential equation are y1 = e-5x and y2 = e6x.
The Wronskian of two functions is given by the determinant of the matrix [y1 y2 y1' y2'], where y1' and y2' are the first derivatives of y1 and y2 respectively. In this case, the Wronskian of e-5x and e6x can be calculated as:
![W[e-5x, e6x] = det | e-5x e6x -5e-5x 6e6x | = (6e6x)(e-5x) - (-5e-5x)(e6x) = 0.](https://img.qammunity.org/2024/formulas/mathematics/college/ly0mz9v1iyomg3c3hgl9et93xr9ni576ix.png)
Since the Wronskian is zero, it indicates that the two functions e-5x and e6x are linearly dependent, which contradicts the given information. Therefore, the statement that the two functions are linearly independent is false.