for what values of r does the function y=e^rx satisfy the differential equation y''-4y'+y=0
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y = 1 e^(rx)
y' = r e^(rx)
y" = r^2 e^(rx)
r^2 - 4 r + 1 = 0
r = [4 +/- sqrt (16 -4) ] /2
r = [ 4 +/- 2 sqrt 3 ]2
r = 2 +/- sqrt 3
To find the values of r for which the function y = e^(rx) satisfies the differential equation y'' - 4y' + y = 0, we need to substitute the function y into the differential equation and solve for r.
Step 1: Find the first and second derivatives of y = e^(rx).
The first derivative of y with respect to x is given by:
y' = (d/dx)(e^(rx)) = r e^(rx)
The second derivative of y with respect to x is given by:
y'' = (d^2/dx^2)(e^(rx)) = r^2 e^(rx)
Step 2: Substitute the derivatives of y into the differential equation.
Substituting y, y', and y'' into the differential equation, we get:
y'' - 4y' + y = r^2 e^(rx) - 4r e^(rx) + e^(rx) = 0
Step 3: Simplify the expression.
To simplify the expression, factor out e^(rx):
e^(rx)(r^2 - 4r + 1) = 0
Step 4: Set each factor equal to zero.
Setting e^(rx) = 0 will not give us any meaningful solutions, so we focus on the quadratic factor:
r^2 - 4r + 1 = 0
Step 5: Solve the quadratic equation.
We can solve the quadratic equation using the quadratic formula:
r = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -4, and c = 1. Substituting these values into the quadratic formula, we get:
r = (-(-4) ± √((-4)^2 - 4(1)(1))) / (2(1))
= (4 ± √(16 - 4)) / 2
= (4 ± √(12)) / 2
= (4 ± 2√(3)) / 2
= 2 ± √(3)
Thus, the solutions for r are:
r1 = 2 + √(3)
r2 = 2 - √(3)
Therefore, the values of r for which the function y = e^(rx) satisfies the differential equation y'' - 4y' + y = 0 are r = 2 + √(3) and r = 2 - √(3).
To find the values of r for which the function y = e^rx satisfies the differential equation y'' - 4y' + y = 0, we need to substitute the function into the differential equation and solve for r.
Step 1: Substitute y = e^rx into the differential equation.
The differential equation is: y'' - 4y' + y = 0
Substituting y = e^rx:
(e^rx)'' - 4(e^rx)' + e^rx = 0
Step 2: Differentiate the function y = e^rx.
Recall that the derivative of e^rx with respect to x is re^rx. Using this, we find the second and first derivatives of e^rx:
(e^rx)'' = r^2e^rx
(e^rx)' = re^rx
Step 3: Substitute the derivatives into the differential equation.
(r^2e^rx) - 4(re^rx) + e^rx = 0
Step 4: Combine like terms.
r^2e^rx - 4re^rx + e^rx = 0
Step 5: Factor out common terms.
e^rx(r^2 - 4r + 1) = 0
Step 6: Set each factor equal to zero and solve for r.
e^rx = 0 (This is not possible because e^rx is always positive)
r^2 - 4r + 1 = 0
Step 7: Solve the quadratic equation.
Using the quadratic formula: r = (-b ± √(b^2 - 4ac)) / (2a)
For the equation r^2 - 4r + 1 = 0, a = 1, b = -4, and c = 1.
r = (-(-4) ± √((-4)^2 - 4(1)(1))) / (2(1))
r = (4 ± √(16 - 4)) / 2
r = (4 ± √12) / 2
r = (4 ± 2√3) / 2
Simplifying further, we get:
r1 = 2 + √3
r2 = 2 - √3
Therefore, the values of r that make the function y = e^rx satisfy the differential equation y'' - 4y' + y = 0 are r1 = 2 + √3 and r2 = 2 - √3.