The general solution to the second-order differential equation y″+6y′−40y=0 is in the form y(x)=c1er1x+c2er2x. Find the values of r1 and r2.

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 () = 0  

Characteristic equation =

                             =

                               = 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

22 and 22

step-by-step explanation:

10 plus 12 equals 22.

12 plus 10 also equals 22 because regardless of what order they're in, they're gonna equal the same thing.

56

step-by-step explanation: