(x^2+1)y'=xy

Jasonbradley

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Jun 17, 2024

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

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

(x² + 1)y' = xy

1. Separating Variables:

The first step is to separate the variables, meaning we get all the y terms on one side of the equation and all the x terms on the other side.

• Divide both sides by y: (x² + 1)y'/y = x
• Divide both sides by (x² + 1): y'/y = x/(x² + 1)

2. Integrating Both Sides:

Now we can integrate both sides of the equation.

• The left side integrates as ln|y|: ∫(y'/y) dy = ln|y| + C₁
• The right side requires a substitution. Let u = x² + 1, then du = 2x dx: ∫(x/(x² + 1)) dx = (1/2)∫(1/u) du = (1/2)ln|u| + C₂ = (1/2)ln|x² + 1| + C₂

3. Combining Constants and Solving for y:

We can combine the constants of integration, C₁ and C₂, into a single constant C:

ln|y| = (1/2)ln|x² + 1| + C

To solve for y, we can exponentiate both sides:

|y| = e^((1/2)ln|x² + 1| + C)

Simplifying the right side:

|y| = e^(ln|x² + 1|^(1/2)) * e^C

|y| = |x² + 1|^(1/2) * e^C

Since e^C is a constant, we can replace it with another constant K:

|y| = K|x² + 1|^(1/2)

Finally, removing the absolute values, we get the general solution:

y = ±K√(x² + 1)

4. Conclusion:

We have successfully found the general solution to the differential equation (x² + 1)y' = xy. The solution is a family of curves defined by the equation y = ±K√(x² + 1), where K is an arbitrary constant.

This solution represents a set of curves that are symmetric about the x-axis and have a shape similar to a hyperbola. The constant K determines the "scale" or size of these curves.