Check that `y_1, y_2, y_3` are solution to the differential equation:
`y'_1=e^x`
`y''_1=e^x`
`y'''_1=e^x`
`y'''+2y''-y'-2y=e^x+2e^x-e^x-2e^x=0`
Therefore `e^x ` is a solution to the differential equation.
Same for `y_2(x)=e^(-x)`
`y'_2(x)=-e^(-x) `
`y''_2(x)=e^(-x) `
`y'''_2(x)=-e^(-x) `
`y'''+2y''-y'-2y=-e^(-x)+2e^(-x)-(-e^(-x)-2e^(-x)=0`
Therefore `e^(-x) ` is a solution to the differential equation.
Same for y`_3(x)=e^(-2x)`
`y'_3(x)=-2e^(-2x)`
`y''_3(x)=4e^(-2x)`
`y'''_3(x)=-8e^(-2x)`
`y'''+2y''-y'-2y=-8e^(-2x)+8e^(-2x)-(-2e^(-2x)-2e^(-2x)=0`
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.`
Let `X=e^x`
`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.
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