Question

Find the particular solution for the initial value problem.

d/y + 6x² = 4x; y(0) = 1 dx
First write the general solution to the differential equation that can be used to find the particular solution.

Answer 1

The particular solution to the differential equation dy/dx + 6x² = 4x with initial condition y(0) = 1 is y = 2x² - 2x³ + 1.

Step-by-step explanation:

Finding a Particular Solution for a Differential Equation with an Initial Condition

To find the particular solution to the initial value problem dy/dx + 6x² = 4x with the initial condition y(0) = 1, we must first find the general solution to the differential equation and then use the given initial condition to determine the particular constants.

The differential equation can be rearranged to isolate the derivative: dy/dx = 4x - 6x². This is a separable equation, and we can integrate both sides to find the general solution:

1. Integrate the right-hand side with respect to x: ∫(4x - 6x²) dx = 4/2 x² - 6/3 x³ = 2x² - 2x³.
2. Now integrate the left-hand side with respect to y, which is simply y.
3. After integration, we equate both integrals and add a constant of integration C: y = 2x² - 2x³ + C.
4. Apply the initial condition y(0) = 1 to find C: 1 = 2(0)² - 2(0)³ + C, hence C = 1.
5. Substitute C back into the general solution to get the particular solution: y = 2x² - 2x³ + 1.

The particular solution of the initial value problem is y = 2x² - 2x³ + 1 where y(0) = 1.