dx%2Fdy%3Dxy-y(-1)%3D1.jpeg "(x^2+2y^2)dx/dy=xy Y(-1)=1")
Solving the Differential Equation: (x^2 + 2y^2) dx/dy = xy, y(-1) = 1
This problem involves solving a first-order, non-linear differential equation with an initial condition. Let's break down the steps to find the solution.
1. Rearranging the Equation
First, we need to separate the variables x and y. Divide both sides by (x^2 + 2y^2) and multiply both sides by dy:
dx = (xy / (x^2 + 2y^2)) dy
2. Integrating Both Sides
Now we integrate both sides of the equation with respect to their respective variables:
∫ dx = ∫ (xy / (x^2 + 2y^2)) dy
The left side is straightforward:
x = ∫ (xy / (x^2 + 2y^2)) dy
The right side requires a substitution. Let u = x^2 + 2y^2, then du = 4y dy. We can rewrite the integral:
x = ∫ (1/4) (u/u) du = (1/4) ∫ du
Integrating and substituting back for u:
x = (1/4)u + C = (1/4)(x^2 + 2y^2) + C
3. Applying the Initial Condition
We are given the initial condition y(-1) = 1. Substitute these values into the equation:
-1 = (1/4)(-1)^2 + (1/2) + C
Solving for C, we get C = -1.
4. The Solution
Substituting the value of C back into the equation, we get the solution to the differential equation:
x = (1/4)(x^2 + 2y^2) - 1
5. Simplifying the Solution (Optional)
We can rearrange the equation to make it more explicit:
x^2 - 4x + 2y^2 + 4 = 0
Conclusion
The solution to the differential equation (x^2 + 2y^2) dx/dy = xy with the initial condition y(-1) = 1 is x^2 - 4x + 2y^2 + 4 = 0. This solution represents a family of curves satisfying the given differential equation and initial condition.