General form of a linear differential equation of order one

$y' + ya(x) = s(x)$

The equation is called homogeneous when $$s(x) = 0$$ otherwise it is called inhomogeneous.

To solve such type of equation we need $$y_h(x)$$ and $$y_p(x)$$.

$y(x) = y_h(x) + y_p(x)$

### $$y_h(x)$$

Where $$y_h(x)$$ is the solution of the homogeneous linear differential equation with the form

$y' + ya(x) = 0$

To get $$y_h(x)$$ we have to separate x and y and calculate the integral for both sides

$\frac{dy}{dx} + ya(x) = 0$ $dy+ya(x)dx = 0$ $\frac{dy}{y} + a(x)dx = 0$

The next step is called separation of variables

$\frac{dy}{y} = -a(x)dx$ $\int \frac{1}{y}dy = - \int a(x)dx$ $ln|y| = - \int a(x)dx + C$ $y = C * e^{- \int a(x)dx}$

for $$C = \pm e^{C_0}$$, $$C_0 \epsilon \R$$ and wee need $$C = 0$$ for $$y = 0$$, because $$e^x \ne 0$$.

Now we can solve $$y_h(x)$$

$y_h(x) = C * e^{- \int a(x)dx}$

### $$y_p(x)$$

To get a particular solution $$y_p(x)$$ we have to change our constant $$C$$ to a function $$C(x)$$. This step is called variation of constants.

$y_p(x) = C(x) * e^{- \int a(x)dx}$

In the original equation we can replace $$y'$$ with $$y_p'(x)$$ and $$y$$ with $$y_p(x)$$.

$y_p'(x) + y_p(x)a(x) = s(x)$ $C'(x) * (e^{- \int a(x)dx}) + C(x) * e^{- \int a(x)dx}*a(x) = s(x)$ $C'(x) * e^{- \int a(x)dx} - C(x)*e^{- \int a(x)dx}*a(x) + C(x) * e^{- \int a(x)dx}*a(x) = s(x)$ $C'(x) = s(x)e^{\int a(x)dx}$ $C(x) = \int s(x)e^{\int a(x)dx}dx$

### $$y(x)$$

Now we can solve our equation

$y(x) = y_h(x) + y_p(x)$ $y(x) = C * e^{- \int a(x)dx} + C(x) * e^{- \int a(x)dx}$