The general solution of the equation y16y=0 is y=c1e4x+c2e4x.F.pdf

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The document presents the solution to a differential equation where y'''-16y=0. The general solution is given as y=c1e^4x + c2e^-4x. The values of c1 and c2 are then determined by using the initial conditions y(0)=4 and y'(0)=-16. By setting up and solving two equations, the unique solution is found to be y=4e^(-4x).

Formation

The general solution of the equation y???16y=0 is y=c1e4x+c2e?4x.
Find values of c1 and c2 so that y(0)=4 and y?(0)=?16.
c1=
c2=
Plug these values into the general solution to obtain the unique solution.
y=
Solution
y = c1*e^(4x) + c2*e^(-4x)
y(0) = 4 :
4 = c1e^0 + c2e^0
c1 + c2 =4 ---> EQUATION ONE
y' = 4c1e^(4x) - 4c2e^(-4x)
y'(0) = -16 :
-16 = 4c1e^0 - 4c2e^0
4c1 - 4c2 = -16
c1 - c2 = -4 ---> EQUATION TWO
c1 + c2 = 4
c1 - c2 = -4
Adding :
2c1 = 0
c1 = 0 --> ANSWER
c2 = 4 --> ANSWER
So, the unique solution is :
y = 0e^(4x) + 4e^(-4x)
y = 4e^(-4x) ---> ANSWER

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