(x^2+2y^2)dx/dy=xy Y(-1)=1

Jasonbradley

3 min read

·

Jun 17, 2024

26

Share

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.