Question

Solve the differential equation, subject to the given initial condition: x(dy/dx) + 3y = 2x^2, y(3) = 12.

a. y = x^3 - 3x^2 + 6x - 3
b. y = x^3 - 3x^2 + 6x + 3
c. y = x^3 - 3x^2 + 3x - 3
d. y = x^3 - 3x^2 + 3x + 3

Answer 1

To solve the differential equation x(dy/dx) + 3y = 2x^2 with y(3) = 12, we use the integrating factor method. After separation of variables and integration, applying the initial condition gives us the particular solution y = x^3 - 3x^2 + 6x + 3.

Step-by-step explanation:

We are given the differential equation x(dy/dx) + 3y = 2x2 with the initial condition y(3) = 12. To solve this equation, let's separate variables and integrate.

1. Divide the entire equation by x to get dy/dx + 3y/x = 2x.
2. This is a first-order linear differential equation, which can be solved using the integrating factor method. Multiply through by the integrating factor, which is e3ln|x| or simply x3.
3. The equation now becomes x3 (dy/dx) + 3x2y = 2x5.
4. Recognizing that the left side of the equation is the derivative of x3y, integrate both sides with respect to x to obtain x3y = (2/6)x6 + C.
5. Divide through by x3 to solve for y, yielding y = (1/3)x3 + C/x3.
6. Apply the initial condition y(3) = 12 to solve for C.
7. After finding C, put it back into the equation to find the particular solution y = x3 - 3x2 + 6x + 3, which corresponds to choice (d).