y′′′−6y′′+ 11y′−6y= 0 wherey(0) =y′(0) = 0, y′′(0) = 2

AAAaannndd the bot gets it wrong yet again!

y = c1 e^x + c2 e^(2x) + c3 e^(3x)
y' = c1 e^x + 2c2 e^(2x) + 3c3 e^(3c)
y" = c1 e^x + 4c2 e^(2x) + 9c3 e^(3x)
The initial conditions give us
c1 + c2 + c3 = 0
c1 + 2c2 + 3c3 = 0
c1 + 4c2 + 9c3 = 2

so c1 = 1, c2 = -2, c3 = 1 and
y(x) = e^x - 2e^(2x) + e^(3x) = e^x (e^x - 1)^2

The solution to this differential equation is y(x) = 2e^x - e^2x - e^3x.

Did you know?

Did you know that the given differential equation represents a special type of equation called a third-order linear homogeneous equation? The equation involves derivatives of a function, represented by y, and describes a relationship between these derivatives and the function itself. In this case, the equation y′′′−6y′′+11y′−6y=0 indicates that the third derivative of y, minus six times the second derivative, plus eleven times the first derivative, and minus six times y itself, equals zero. The initial conditions, y(0) = y′(0) = 0, y′′(0) = 2, provide starting values for the function and its derivatives. Solving this equation can reveal insights into various topics such as vibrations, oscillations, and resonance in physics and engineering.