Solve the following initial value problem:
y=∫(3x2−6)dx, y(1)=2
y=
Accepted Solution
A:
To find the value of y for the given integral and initial condition, let's first evaluate the integral:
∫(3x^2 - 6) dx
To integrate the function 3x^2 - 6, we apply the power rule of integration:
∫(3x^2 - 6) dx = x^3 - 6x + C
Now, we can find the value of the constant C by using the initial condition y(1) = 2. Substituting x = 1 into the expression for y, we have:
2 = (1)^3 - 6(1) + C
Simplifying:
2 = 1 - 6 + C
2 = -5 + C
To isolate C, we add 5 to both sides:
C = 2 + 5
C = 7
Now that we have found the value of C, we can rewrite the integral with the specific constant:
∫(3x^2 - 6) dx = x^3 - 6x + 7
Thus, the value of y for the given integral and initial condition is y = x^3 - 6x + 7.