Question

Xydx + (1 + y²)√1 + x² dy = 0​

Answer 1

Ответ:

$\frac{1}{2} y^2+ln|y|=-\sqrt{1+x^2} +C.$

Пошаговое объяснение:

Answer 2

The equation you provided is a first-order nonlinear ordinary differential equation. It appears to be in a standard form:

Xydx + (1 + y²)√(1 + x²) dy = 0

To solve this equation, we can attempt to separate the variables and integrate both sides. Let's go through the steps:

1. Rearrange the equation to have dy/dx terms on one side and the other terms on the other side:

(1 + y²)√(1 + x²) dy = -Xydx

2. Divide both sides by (1 + y²)√(1 + x²) to isolate dy/dx:

dy / (1 + y²) = -Xydx / (1 + y²)√(1 + x²)

3. Integrate both sides with respect to their respective variables:

∫(dy / (1 + y²)) = ∫(-Xydx / (1 + y²)√(1 + x²))

4. Evaluate the integrals:

Let's solve each integral separately:

∫(dy / (1 + y²)) = arctan(y) + C1

∫(-Xydx / (1 + y²)√(1 + x²)) = -X∫(ydx / (1 + y²)√(1 + x²))

This integral may not have a simple closed-form solution. If you have any specific values for X, we can try approximating the integral using numerical methods like numerical integration or numerical solvers.

5. Combine the results:

arctan(y) + C1 = -X∫(ydx / (1 + y²)√(1 + x²))

Therefore, the solution to the given differential equation is given by:

arctan(y) + C1 = -X∫(ydx / (1 + y²)√(1 + x²))

where C1 is the constant of integration.