Answer:
r1 = 4 and r2 = -10 and the final equation will be C1e^4x + C2e^-10x
Step-by-step explanation:
The general solution to the second order differential equation
y'' + 6y' -40y=0
substitute y= e^rx
y' = re^rx
y'' = r^2*e^rx
The equation will be
r^2*e^rx + 6r*e^rx - 40*e^rx = 0
e^rx ([tex]\\r^2 + 6r - 40[/tex]) = 0
Characteristic equation = [tex]\\r^2 + 6r - 40[/tex]
= [tex]r^2 +10r - 4r -40\\[/tex]
= r(r+10) - 4(r+10)
= (r-4)(r+10)
r1 = 4 and r2= -10
y = e^4x and y = e^-10x
the equation will be C1e^4x + C2e^-10x
