Final answer:
To solve the initial value problem yy′ = x − y² with the initial condition y(0) = -6, we can use separation of variables. The general solution is (1/2)y² + tan^(-1)(y) = C + 1/2x², where C is the constant of integration. Substituting the initial condition gives the particular solution.
Step-by-step explanation:
To solve the initial value problem yy′ = x − y², where y(0) = −6, we can use separation of variables. First, let's rearrange the equation to have y' on one side and y and x on the other side. This gives us yy′ + y² = x. Now, we can separate the variables by dividing both sides by (y² + 1): (yy′ + y²)/(y² + 1) = x/(y² + 1).
Next, we can integrate both sides of the equation with respect to y. This will give us the general solution: ∫(yy′ + y²)/(y² + 1) dy = ∫x/(y² + 1) dy. After integrating, we get (1/2)y² + tan^(-1)(y) = C + 1/2x², where C is the constant of integration.
To find the particular solution that satisfies the initial condition y(0) = -6, we substitute y = -6 and x = 0 into the equation. This gives us (1/2)(-6)² + tan^(-1)(-6) = C + 0. Solving for C, we find that C = -18 + tan^(-1)(-6).
Therefore, the solution to the initial value problem yy′ = x − y², with the initial condition y(0) = -6, is (1/2)y² + tan^(-1)(y) = -18 + tan^(-1)(-6) + 1/2x².