Find f (x) by solving the initial value problem.
f '(x) = 6x^2 - 10x; f(2) = 5
f(x) =
I am only going to do one of these because if you have seen one you have seen them all.
f(x) = 6 (x^3/3) - 10 (x^2/2) + c
= 2 x^3 - 5 x^2 + c
no find c by using the one point given
5 = 2 (2)^3 - 5 (2)^2 + c
5 = 16 - 20 + c
c = 9
so
f(x) = 2 x^3 - 5 x^2 + 9
To find the function f(x) by solving the initial value problem, we'll integrate the given differential equation and then use the initial condition to find the constant of integration.
Step 1: Integrate the differential equation
To find f(x), we'll integrate the given differential equation, f'(x) = 6x^2 - 10x, with respect to x. The integral of 6x^2 - 10x is:
∫ (6x^2 - 10x) dx
= 2x^3 - 5x^2 + C
Here, C is the constant of integration.
Step 2: Apply the initial condition
We're also given the initial condition f(2) = 5, which means that when x = 2, f(x) = 5. We can use this information to solve for the constant of integration, C.
Plugging in x = 2 and f(x) = 5 into the integrated equation, we get:
2(2)^3 - 5(2)^2 + C = 5
16 - 20 + C = 5
C - 4 = 5
C = 9
Step 3: Write the final solution
Now that we have found the constant of integration, we can write the final solution for f(x):
f(x) = 2x^3 - 5x^2 + 9
So, the function f(x) that satisfies the differential equation f'(x) = 6x^2 - 10x and the initial condition f(2) = 5 is given by f(x) = 2x^3 - 5x^2 + 9.