(1+x^2)y'-2xy=0

Jasonbradley

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Jun 16, 2024

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Solving the Differential Equation (1+x^2)y' - 2xy = 0

This article will explore the solution to the first-order differential equation:

(1 + x^2)y' - 2xy = 0

We will use methods of separation of variables and integration to find the general solution.

1. Rearranging the Equation

First, we need to rearrange the equation to separate the variables:

(1 + x^2)y' = 2xy

Now, we can divide both sides by (1 + x^2)y to get:

y'/y = 2x/(1 + x^2)

2. Integrating Both Sides

The left-hand side can be rewritten as d(ln|y|)/dx, while the right-hand side requires a substitution.

Let u = 1 + x^2, then du = 2x dx.

The equation becomes:

d(ln|y|)/dx = du/u

Now, we can integrate both sides with respect to x:

∫d(ln|y|)/dx dx = ∫du/u

This simplifies to:

ln|y| = ln|u| + C

Where C is the constant of integration.

3. Solving for y

We can solve for y by taking the exponential of both sides:

|y| = e^(ln|u| + C)

Using the properties of exponentials:

|y| = e^(ln|u|) * e^C

Simplifying further:

|y| = |u| * C₁

Where C₁ = e^C is a new constant.

Since we are dealing with absolute values, we can remove them by introducing a sign:

y = ± C₁(1 + x^2)

4. General Solution

Finally, we can combine the constants into a single arbitrary constant, K:

y = K(1 + x^2)

This is the general solution to the differential equation (1 + x^2)y' - 2xy = 0.

5. Conclusion

We have successfully solved the differential equation using separation of variables and integration. The general solution, y = K(1 + x^2), represents a family of curves that satisfy the original equation.