For which values of r does the function defined by y=e^(rt) satisfy the differential equation y''+ y'-6y = 0
To find the values of r for which the function defined by y=e^(rt) satisfies the differential equation y''+ y'-6y = 0, we need to substitute the function into the equation and solve for r.
First, let's differentiate y = e^(rt) twice.
y' = r*e^(rt)
y'' = r^2*e^(rt)
Now, substitute these expressions into the differential equation:
r^2*e^(rt) + r*e^(rt) - 6*e^(rt) = 0
Since e^(rt) is a nonzero function for any value of t, we can divide the entire equation by e^(rt):
r^2 + r - 6 = 0
Now, we can factorize the quadratic equation:
(r + 3)(r - 2) = 0
Setting each factor equal to zero, we have two possible values for r:
r + 3 = 0 => r = -3
r - 2 = 0 => r = 2
Therefore, the values of r for which the function y = e^(rt) satisfies the differential equation y''+ y'-6y = 0 are r = -3 and r = 2.
To find the values of r for which the function y = e^(rt) satisfies the differential equation y'' + y' - 6y = 0, we need to substitute this function into the differential equation and solve for r.
First, let's find the first and second derivatives of y = e^(rt):
y' = r * e^(rt)
y'' = r^2 * e^(rt)
Substituting these derivatives into the differential equation:
r^2 * e^(rt) + r * e^(rt) - 6 * e^(rt) = 0
Now, we can factor out e^(rt) from the equation:
e^(rt) * (r^2 + r - 6) = 0
The exponential function e^(rt) is never zero, so we can divide both sides of the equation by e^(rt):
r^2 + r - 6 = 0
Now, we can factor the quadratic equation:
(r + 3)(r - 2) = 0
Setting each factor equal to zero gives two possible solutions:
r + 3 = 0 --> r = -3
r - 2 = 0 --> r = 2
Therefore, the values of r that make the function y = e^(rt) satisfy the differential equation y'' + y' - 6y = 0 are r = -3 and r = 2.