Question

Substitute y=e^rx into the given differentialequation to determine all values of the constant r for which y=e^rx is a solution of the equation.3y''+3y'-4y=0

Answer 1

To determine the values of the constant r for which y = e^rx is a solution of the given differential equation, we substitute y = e^rx into the equation and solve for r. The values of r that make y = e^rx a solution are approximately (-3 + √57)/6 and (-3 - √57)/6.

Step-by-step explanation:

To determine the values of the constant r for which the function y = e^rx is a solution of the given second order linear differential equation 3y''+3y'-4y=0, we substitute y = e^rx into the equation and solve for r.

Substituting y = e^rx into the equation, we obtain:

3(e^rx)'' + 3(e^rx)' - 4(e^rx) = 0

Simplifying and rearranging the equation, we get:

3r^2e^rx + 3re^rx - 4e^rx = 0

Factoring out the common factor e^rx, we have:

(3r^2 + 3r - 4)e^rx = 0

For the equation to hold true for all values of x, the coefficient of e^rx must be equal to zero.

Therefore, we solve the quadratic equation 3r^2 + 3r - 4 = 0 to find the values of r that make y = e^rx a solution.

Using the quadratic formula, we have:

r = (-3 ± √(3^2 - 4(3)(-4))) / (2(3))

Simplifying further, we get:

r = (-3 ± √(9 + 48)) / 6

r = (-3 ± √57) / 6

So, the values of r for which y = e^rx is a solution of the equation are approximately:

r ≈ (-3 + √57) / 6

r ≈ (-3 - √57) / 6

Answer 2

`r = (-3 - √(57))/(6),(-3 + √(57))/(6)`

Step-by-step explanation:

We are given the following in the question:

`y = e^(rx)`

It is a solution to the differential equation

`3y''+3y'-4y=0`

`y' = (dy)/(dx) = (d(e^(rx)))/(dx) = re^(rx)\\\\y'' = (d(y'))/(dx) = (d(re^(rx)))/(dx) = r^2e^(rx)`

Putting the values in the differential equation we get,

`3(r^2e^(rx)) + 3(re^(rx))-4e^(rx) = 0\\(3r^2+3r-4)e^(rx) = 0\\e^(rx)\\eq 0\\3r^2+3r-4 = 0\\\\r = (-3\pm √(9-4(3)(-4)))/(6)\\\\r =(-3\pm √(57))/(6)\\\\r =(-3 - √(57))/(6),(-3 + √(57))/(6)`