Differential Equation - Exercise 20

We are tasked with solving the differential equation:

$$ y' + 2x y = 0 $$

This equation can be written in separable form $ y' = a(x) \, b(y) $, where $ a(x) = -2x $ and $ b(y) = y $:

$$ y' = - 2x y $$

Express the derivative explicitly as $ \frac{dy}{dx} $:

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

Separate the variables:

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

Integrate both sides:

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

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

The integral of $ x $ is $ \frac{x^2}{2} + C_1 $:

$$ \int \frac{1}{y} \, dy = -2 \left( \frac{x^2}{2} + C_1 \right) $$

$$ \int \frac{1}{y} \, dy = - x^2 - 2C_1 $$

Since $-2C_1$ is still an arbitrary constant, we can rename it as $ C_2 $:

$$ \int \frac{1}{y} \, dy = - x^2 + C_2 $$

The antiderivative of $ 1/y $ is $ \log(y) $:

$$ \log(y) = - x^2 + C_2 $$

Exponentiate both sides to isolate $ y $:

$$ e^{\log(y)} = e^{-x^2 + C_2} $$

$$ y = e^{-x^2} \cdot e^{C_2} $$

Since $ e^{C_2} $ is just a constant, we can write:

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

This is the general solution of the equation.

Constant solution: In addition to the general family of solutions, there is also a constant solution. Setting $ y = 0 $ gives $ y' = 0 $, which satisfies the original equation: $$ y' + 2xy = 0 $$ $$ (0) + 2x(0) = 0 $$ $$ 0 = 0 $$ Thus, $ y \equiv 0 $ is also a valid solution.