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Sagot :
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.
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