Differential Equation Exercise 17

We want to solve the following first-order differential equation:

$$ xy' + 2y = x $$

Applying the invariance property, we divide through by $ x $:

$$ \frac{xy' + 2y}{x} = \frac{x}{x} $$

$$ \frac{xy'}{x} + \frac{2y}{x} = 1 $$

$$ y' + \frac{2y}{x} = 1 $$

This is now in the form of a linear differential equation:

$ y' + a(x)\,y = b(x) $, where $ a(x) = \frac{2}{x} $ and $ b(x) = 1 $.

We can solve this using the method of variation of constants:

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

Substituting $ a(x) = \frac{2}{x} $ and $ b(x) = 1 $:

$$ y = e^{-\int \frac{2}{x} \, dx} \left[ \int e^{\int \frac{2}{x} \, dx} \, dx + c \right] $$

$$ y = e^{-2 \int \frac{1}{x} \, dx} \left[ \int e^{2 \int \frac{1}{x} \, dx} \, dx + c \right] $$

Since $ \int \frac{1}{x} \, dx = \log x $, this becomes:

$$ y = e^{-2 \log x} \left[ \int e^{2 \log x} \, dx + c \right] $$

Using $ e^{\log x} = x $:

$$ y = x^{-2} \left[ \int x^2 \, dx + c \right] $$

The integral $ \int x^2 \, dx $ is $ \frac{x^3}{3} $:

$$ y = x^{-2} \left[ \frac{x^3}{3} + c \right] $$

Distributing $ x^{-2} $:

$$ y = \frac{x}{3} + c \, x^{-2} $$

This is the general solution of the differential equation.