Verify that y1(x)=e^x, y2(x)=e^-x, y3(x)=e^-2x form a linearly independent set of solution of y''' + 2y'' - y' - 2y =0
Check that `y_1, y_2, y_3` are solution to the differential equation:
Therefore `e^x ` is a solution to the differential equation.
Same for `y_2(x)=e^(-x)`
Therefore `e^(-x) ` is a solution to the differential equation.
Same for y`_3(x)=e^(-2x)`
Therefore `e^(-2x)` is a solution to the differential equation.
`y_1, y_2, y_3` are 3 solutions to the differential equation.
Let's prove that they are linearly independent.
Let a,b,c such that `ay_1+by_2+cy_3=0`
`AA x inRR, ae^x+be^(-x)+ce^(-2x)=0 `
Multiply the equality by `e^(2x)`
`AAx inRR, ae^(3x)+be^(x)+c=0.`
`AA X inRR^+, aX^3+bX+c=0. `
It is a polynomial function with infinitely many roots therefore all its coefficients are 0.
a=0, b=0, c=0 and the 3 functions are linearly independent.