Determine whether the function y = 2x³ - 3x + 3 is a solution to the initial-value problem (IVP): y' = 6x² - 3 and y(1) = 2.



Answer :

Yes, the function y = 2x³ - 3x + 3 is a solution to the IVP: y' =  6x² - 3 and y(1) = 2.

Work

To determine if the function is a solution to the initial value problem, we need to check if the derivate of the function matches up with that of the IVP.

Taking the derivative of: y = 2x³ - 3x + 3 with respect to x

y' = 6x² - 3


The derivatives are indeed the same. Now, we need check the initial condition by substituting the values of y(1) = 2 into the original function.

y = 2x³ - 3x + 3

2 = 2(1)³ - 3(1) + 3

2 = 2 - 3 + 3

2 = 2

Therefore, the function y = 2x³ - 3x + 3 is indeed a solution to the IVP y' = 6x² - 3 as both conditions are satisfied.

The function y = 2x³ - 3x + 3 is a solution to the initial-value problem y' = 6x² - 3 and y(1) = 2, since it satisfies both the differential equation and the initial condition.

To determine if the function y = 2x³ - 3x + 3 is a solution to the initial-value problem (IVP) given by y' = 6x² - 3 and y(1) = 2, follow these steps:

Step 1: Compute the Derivative

To check if the function satisfies the differential equation y' = 6x² - 3, first find the derivative of the function y:

  • Given: y = 2x³ - 3x + 3
  • The derivative of y is computed as follows:
  • y' = d/dx (2x³ - 3x + 3)
  • Applying the power rule: y' = 6x² - 3

The result matches the differential equation y' = 6x² - 3.

Step 2: Verify the Initial Condition

Next, check if the function satisfies the initial condition y(1) = 2:

  • Substitute x = 1 into y = 2x³ - 3x + 3:
  • y(1) = 2(1)³ - 3(1) + 3 = 2 - 3 + 3 = 2

The result matches the initial condition y(1) = 2.

Since the function y = 2x³ - 3x + 3 satisfies both the differential equation y' = 6x² - 3 and the initial condition y(1) = 2, it is a solution to the initial-value problem.