Differential Equation - Exercise 5

We want to solve the differential equation:

$$ y' + x y = 0 $$

This is a first-order differential equation because the highest derivative of the unknown function is the first derivative $ y' $.

We can solve it in two equivalent ways.

Method 1 - Separation of Variables

We can solve the equation using the separation of variables technique.

Starting with:

$$ y' + x y = 0 $$

Rewrite $ y' $ as $ \frac{dy}{dx} $:

$$ \frac{dy}{dx} + x y = 0 $$

Separate the variables $ x $ and $ y $:

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

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

Integrate both sides:

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

$$ \ln|y| = -\frac{x^2}{2} + C $$

Exponentiate both sides to isolate $ y $:

$$ y = e^{-\frac{x^2}{2} + C} $$

Since $ e^C $ is just a constant, we can write:

$$ y = C \, e^{-\frac{x^2}{2}} $$

This is the general solution to the equation.

The equation is of the standard first-order linear form $ y' + a(x) y = b(x) $, with $ a(x) = x $ and $ b(x) = 0 $:

$$ y' + x y = 0 $$

The general solution is given by:

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

Substitute $ a(x) = x $ and $ b(x) = 0 $:

$$ y = e^{-\int x \, dx} \cdot C $$

Since $ \int x \, dx = \frac{x^2}{2} $, we get:

$$ y = C \, e^{-\frac{x^2}{2}} $$

Thus, the general solution is:

$$ y = C \, e^{-\frac{x^2}{2}} $$

Both methods lead to exactly the same result, confirming its correctness.