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Sagot :
The general solution of the given higher-order differential equation y''' − 8y'' − 9y' = 0 is , [tex]y = C_{1} + C_{2} e^{8x} + C_{3}e^{-x}[/tex]
Given higher-order differential equation = y''' − 8y'' − 9y' = 0
We have to find the general solution of the above differential equation
The auxiliary equation for the above differential equation will be : [tex]m^{3} - 8m^{2} - 9m = 0[/tex]
Solve the equation :- [tex]m^{3} - 8m^{2} -9m = 0[/tex]
[tex]m ( m^{2} - 8m - 9 ) = 0[/tex]
m ( m - 8 ) ( m + 1 ) = 0
m = 0 , m = 8 , m = -1
Then the general solution will be like :
[tex]y = C_{1} e^{0x} + C_{2} e^{x} +C_{3}e^{-1x}[/tex]
The above solution can be written as:
[tex]y = C_{1} +C_{2} e^{8x} +C_{3} e^{-x}[/tex]
Learn more about differential equation here : https://brainly.com/question/13257443
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