Differential Equation - Exercise 9

We are tasked with solving the following first-order differential equation:

$$ y' + y = 0 $$

This is a homogeneous linear equation of the first order, and it can be solved using two different standard techniques.

Method 1: Separation of Variables

The equation is of the form y' + f(x)g(y) = 0, where \( f(x) = 1 \) and \( g(y) = y \). Since the variables can be separated, we can apply the separation of variables method.

We begin by isolating the derivative:

$$ y' = -y $$

Rewriting in differential form:

$$ \frac{dy}{dx} = -y $$

Now, we separate the variables:

$$ \frac{dy}{y} = -dx $$

Integrating both sides:

$$ \int \frac{1}{y} \, dy = \int -1 \, dx $$

Evaluating the integrals gives:

$$ \log y = -x + c $$

To solve for \( y \), we exponentiate both sides:

$$ e^{\log y} = e^{-x + c} $$

$$ y = e^{-x} \cdot e^c $$

Since \( e^c \) is a positive constant, we can absorb it into a new constant \( c \):

$$ y = c \cdot e^{-x} $$

This is the general solution to the differential equation.

Method 2: Linear First-Order Equation

The same equation can be viewed as a linear first-order equation of the form y' + a(x)y = b(x), where \( a(x) = 1 \) and \( b(x) = 0 \).

In this case, we apply the integrating factor method (also known as Lagrange’s method). The general solution is given by:

$$ y = e^{-\int a(x) \, dx} \left[ \int b(x) \cdot e^{\int a(x) \, dx} \, dx + c \right] $$

Substituting \( a(x) = 1 \) and \( b(x) = 0 \):

$$ y = e^{-\int 1 \, dx} \left[ \int 0 \cdot e^{\int 1 \, dx} \, dx + c \right] $$

$$ y = e^{-x} \cdot c $$

Once again, we obtain the general solution:

$$ y = c \cdot e^{-x} $$

As expected, both methods yield the same result.

And so on.