To solve the differential equation

$(1+y^2)dx + (1+x^2)dy = 0,$

we start by rearranging it into a more standard form for solving:

$(1+y^2)dx = -(1+x^2)dy.$

This can be written as:

$\frac{dx}{1+x^2} = -\frac{dy}{1+y^2}.$

Now, we can integrate both sides separately:

1. Integrate the left side with respect to $x$:

$\int \frac{dx}{1+x^2} = \tan^{-1}(x) + C_1.$

2. Integrate the right side with respect to $y$:

$\int \frac{dy}{1+y^2} = \tan^{-1}(y) + C_2.$

Since these integrals are indefinite, we combine the constants $C_1$ and $C_2$ into a single constant $C$:

$\tan^{-1}(x) = -\tan^{-1}(y) + C.$

Rewriting the equation to isolate $C$, we get:

$\tan^{-1}(x) + \tan^{-1}(y) = C.$

This is the general solution to the differential equation.

Would you like more details on any part of this solution or have any questions?

Related Questions

1. What are the properties of the inverse tangent function $\tan^{-1}(x)$?
2. How do you solve a differential equation of the form $M(x, y)dx + N(x, y)dy = 0$ using exact equations?
3. Can you always separate variables in a differential equation? Why or why not?
4. What are the steps to verify if a differential equation is exact?
5. How do you integrate trigonometric functions?
6. What is the significance of the constant of integration in solving differential equations?
7. How can you apply initial conditions to find a particular solution of a differential equation?
8. What are some common techniques for solving first-order differential equations?

Tip

When dealing with integrals involving inverse trigonometric functions, remember their derivatives and the typical forms they appear in to quickly recognize and solve them.