Differential Equation Exercise 7

In this exercise, we aim to solve the differential equation:

$$ y' + 2x = 0 $$

This is a first-order differential equation, as the highest derivative involved is the first derivative \( y' \).

The equation can be expressed in the form \( y' = f(x) \cdot g(y) \), where \( f(x) = -2x \) and \( g(y) = 1 \).

This makes it suitable for the separation of variables method.

We begin by isolating \( y' \):

$$ y' = -2x $$

Next, we rewrite the derivative in differential form:

$$ \frac{dy}{dx} = -2x $$

Separating the variables gives:

$$ dy = -2x \, dx $$

We now integrate both sides with respect to their respective variables:

$$ \int dy = \int -2x \, dx $$

The integral on the right-hand side yields the antiderivative \( -x^2 + c \):

$$ y = -x^2 + c $$

Since the left-hand side integrates to \( y \), and the constant of integration is already included on the right, this is our general solution:

$$ y = -x^2 + c $$

Thus, we’ve found the general solution to the differential equation.

And so on.