For what values of r does the function
y = erx
Satisfy the equation
y'' + 5y' + 6y = 0?
do you mean y = e^rx ?
r^2 e^rx + 5 r e^rx + 6 e^rx = 0
r%2 + 5 r + 6 = 0
(r+3)(r+2) = 0
r = -3 or -2
Well, well, well... Look who's dealing with some exponential functions! Now, let's get down to business.
First things first, let's find the derivatives of y = erx.
The first derivative, y' = rex, is just the product of erx and rex.
The second derivative, y'' = rex * rex + erx, is the sum of the product erx * rex and erx itself.
Now, let's plug these babies into the given equation: y'' + 5y' + 6y = 0.
Substituting the derivatives, we get (rex * rex + erx) + 5(rex) + 6(erx) = 0.
Now, let's factor out erx: erx(rex + 5 + 6) = 0.
To satisfy the equation, we need the expression inside the parentheses to be equal to zero.
So, rex + 11 = 0.
Well, we can't have a negative exponential base (unless we want to invite imaginary numbers to the party). So, rex must be equal to zero. And when the base is zero, the exponent can be any real number!
Therefore, the function y = erx satisfies the equation y'' + 5y' + 6y = 0 for all values of r.
Now, wasn't that an exponential-ly funny answer? Ha!
To find the values of r for which the function y = erx satisfies the equation y'' + 5y' + 6y = 0, we need to substitute y = erx into the differential equation and solve for r.
Given the function y = erx, we can find the first and second derivatives of y as follows:
y' = d(erx)/dx = r(erx) = r * erx
y'' = d/dx(r * erx) = r * d(erx)/dx = r(erx) = r^2 * erx
Substituting y = erx into the differential equation, we get:
y'' + 5y' + 6y = 0
r^2 * erx + 5r * erx + 6erx = 0
Factoring out the common factor erx from each term, we have:
erx(r^2 + 5r + 6) = 0
For the equation to be satisfied, either erx = 0 or (r^2 + 5r + 6) = 0.
Since erx is exponential and nonzero for all real values of x, we focus on the quadratic equation (r^2 + 5r + 6) = 0.
Simplifying this quadratic equation, we have:
(r + 2)(r + 3) = 0
Setting each factor equal to zero, we get two possible solutions:
r + 2 = 0 --> r = -2
r + 3 = 0 --> r = -3
Therefore, the values of r for which the function satisfies the equation y'' + 5y' + 6y = 0 are r = -2 and r = -3.
To determine the values of r for which the function y = erx satisfies the given differential equation y'' + 5y' + 6y = 0, we need to substitute y into the differential equation and find the values of r that make the equation true.
Let's start by finding the first derivative of y (y') and the second derivative of y (y'').
First derivative:
y' = d/dx(erx) = rerx
Second derivative:
y'' = d/dx(rerx) = re^rx * r = r^2erx
Now, let's substitute these derivatives into the differential equation:
r^2erx + 5rerx + 6erx = 0
erx can be canceled out since it is common to all terms:
r^2 + 5r + 6 = 0
This is a quadratic equation. To solve it, we can factorize or use the quadratic formula:
(r + 2)(r + 3) = 0
From this equation, we can see that r = -2 or r = -3. These are the values for which the original function y = erx satisfies the given differential equation y'' + 5y' + 6y = 0.