For what values of r does the function
y = erx
satisfy the differential equation
y'' − 6y' + 3y = 0?
To determine the values of r that satisfy the given differential equation, we need to substitute the given function y = e^(rx) into the differential equation and solve for r.
First, let's find the first and second derivatives of y = e^(rx):
y' = r * e^(rx) (using the chain rule)
y'' = r^2 * e^(rx) (using the chain rule)
Now, substitute these derivatives and the original function y into the differential equation:
y'' - 6y' + 3y = 0
(r^2 * e^(rx)) - 6(r * e^(rx)) + 3(e^(rx)) = 0
Factor out e^(rx) from the equation:
e^(rx) * (r^2 - 6r + 3) = 0
For this equation to hold true, either e^(rx) = 0 or (r^2 - 6r + 3) = 0. However, e^(rx) is always positive, so it can never be zero. Therefore, we only need to solve the quadratic equation (r^2 - 6r + 3) = 0.
Using the quadratic formula, we have:
r = (-(-6) ± √((-6)^2 - 4*(1)*(3))) / (2*(1))
r = (6 ± √(36 - 12)) / 2
r = (6 ± √(24)) / 2
r = (6 ± 2√6) / 2
Now, simplify the expression further:
r = 3 ± √6
Therefore, the values of r that satisfy the given differential equation y'' - 6y' + 3y = 0 are r = 3 + √6 and r = 3 - √6.
To solve the differential equation y'' − 6y' + 3y = 0 with the function y = erx, we need to find the values of r for which the function satisfies the equation. Let's start by finding the derivatives of y.
First derivative:
y' = d/dx(erx)
= r * erx
Second derivative:
y'' = d/dx(r * erx)
= r * d/dx(erx)
= r * r * erx
= r^2 * erx
Now, substitute these derivatives into the differential equation:
r^2 * erx − 6r * erx + 3 * erx = 0
Since erx is common in all terms, we can factor it out:
erx * (r^2 - 6r + 3) = 0
For this equation to hold true, either erx = 0 or (r^2 - 6r + 3) = 0.
First case: erx = 0
Since erx is an exponential function, it is never equal to zero for any value of x. Therefore, this case is not possible.
Second case: r^2 - 6r + 3 = 0
To solve this quadratic equation, we can use the quadratic formula:
r = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 1, b = -6, and c = 3.
Plugging in the values, we get:
r = (6 ± √((-6)^2 - 4 * 1 * 3)) / (2 * 1)
r = (6 ± √(36 - 12)) / 2
r = (6 ± √24) / 2
r = (6 ± 2√6) / 2
r = 3 ± √6
Therefore, the values of r for which the function y = erx satisfies the differential equation y'' − 6y' + 3y = 0, are r = 3 + √6 and r = 3 - √6.
the roots of
D^2-6D+3 are -3±√6