Variation of parameters

Consider the equation, where $P,Q$ are functions of $x$ alone, and which has 2 fundamental solutions $y_1,y_2$:

$$y''+Py'+Qy=f(x)$$

The general solution of the equation is:

$$y_g=c_1y_1 + c_2y_2$$

Now replacing $c_1,c_2$ with $u(x),v(x)$ will give $y_p=uy_1 + vy_2$, which can be found using the method of variation of parameters.

$$u=-\int{\frac{y_2f}{W(x)}\,\text{d}x} \;\land\; v=\int{\frac{y_1f}{W(x)}\,\text{d}x}$$

Proof

$$y_p=uy_1 + vy_2$$ $$y_p'=u'y_1 + uy_1' + v'y_2 + vy_2'$$

Set $u'y_1+v'y_2=0\;\;(1)$ to simplify further equations. That implies $y_p'=uy_1'+vy_2'$.

$$y_p''=uy_1''+u'y_1'+vy_2''+v'y_2$$

Substituting $y_p'',y_p',y_p$ to the differential equation: $$

$$y_p''+Py_p'+Qy_p=u'y_1'+v'y_2'$$

This implies $u'y_1'+v'y_2'=f(x)\;\;(2)$. $$

From equations $(1)$ and $(2)$, where $W(x)$ is the wronskian of $y_1,y_2$:

$$u'=-\frac{y_2f}{W(x)} \;\land\; v'=\frac{y_1f}{W(x)}$$ $$u=-\int{\frac{y_2f}{W(x)}\,\text{d}x} \;\land\; v=\int{\frac{y_1f}{W(x)}\,\text{d}x}$$

$y_p$ can be found now using $u,v,y_1,y_2$