Differential Equation Exercise

We want to solve the differential equation

$$ y' = - \frac{1+y}{x^2} $$

This is a first-order differential equation, and it can be handled using the method of separation of variables.

We begin by separating the variables $y$ and $x$:

$$ \frac{1}{1+y}\, y' = - \frac{1}{x^2} $$

Writing $y'$ as dy/dx and rearranging gives

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

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

We now integrate both sides with respect to their variables:

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

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

Since $\int x^{-2}\, dx = \frac{x^{-1}}{-1} = -\frac{1}{x}$, we obtain

$$ \int \frac{1}{1+y}\, dy = \frac{1}{x} + c_1 $$

On the left-hand side, the antiderivative is the natural logarithm:

$$ \log |1+y| = \frac{1}{x} + c_1 $$

Exponentiating both sides yields

$$ 1+y = e^{\frac{1}{x}} \cdot e^{c_1} $$

$$ y = c\, e^{\frac{1}{x}} - 1 $$

where $c = e^{c_1}$ is an arbitrary constant. This is the general solution.

Returning to the original equation,

$$ y' = - \frac{1+y}{x^2}, $$

if $y = -1$ we obtain

$$ y' = -\frac{1+(-1)}{x^2} = 0. $$

Thus $y = -1$ is a valid constant solution, since the derivative vanishes for all $x \neq 0$.

$$ \forall\, x \in \mathbb{R}\setminus \{0\}, \quad y=-1 \implies y'=0 $$

The complete solution set therefore consists of: