Differential Equations.

Question

let y''+6y'-16y=0

a) Try a solution of the form y=e^(rx), for some unknown constant r, by substituting it into the differential equation.
...................=0

b)Simplify and factor your equation in the previous part as much as possible. Be sure to divide out any terms guaranteed to be nonzero.
.........................=0

c)Find all values of r such that y = e^{rx} satisfies the differential equation. If there is more than one correct answer, enter your answers as a comma separated list.
r=............

Answer 1

This is a linear ODE with constant coefficients. A usual approach is to look its solution(s) in the form y=e^rx (as you already mentioned. I guess you know why)

Basically you have all the "instructions" what to do.

1) Since y=e^rx, then y'=re^rx, y''=r²e^rx.  Substitute into equation

y''+6y'-16y=0

r²e^rx+6re^rx-16e^rx=0  (divide by e^rx)

r²+6r-16=0

Solve for r and substitute in y=e^rx.

Continue, if questions, ask again.

 

Answer 2

If y = e^rx, then y' = re^rx and y" = r²e^rx

 

Substituting into the given differential equation, we have (r² + 6r - 16)e^rx = 0

 

    Since e^rx is always positive, we get r² + 6r - 16 = 0

 

                                                       (r+8)(r-2) = 0   So, r = -8 or r = 2

 

General solution of the differential equation is:  y = c₁e^-8x + c₂e^2x