# 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

cosinusix | College Teacher | (Level 3) Assistant Educator

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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.