Differential Equation Exercise 6

We are tasked with solving the following differential equation:

$$ y' + 1 = 0 $$

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

It is a straightforward, elementary case.

Solving it simply requires evaluating an indefinite integral.

First, we isolate the derivative:

$$ y' = -1 $$

We now integrate both sides of the equation with respect to \( x \):

$$ \int y'\, dx = \int -1\, dx $$

$$ \int y'\, dx = - \int 1\, dx $$

The integral on the left yields the antiderivative \( y + c \):

$$ y + c = - \int 1\, dx $$

The integral on the right evaluates to \( x + c \), giving:

$$ y + c = -x $$

To simplify notation, we absorb all constants into a single arbitrary constant \( c \):

Therefore, the general solution is:

$$ y = -x + c $$

Note: We express the arbitrary constant only once, since any combination of constants from both sides can be consolidated. For instance, the expression $$ y + c_1 = -x - c_2 $$ reduces to the same general solution by defining \( c = c_1 + c_2 \): $$ y = -x + c $$

And so on.