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

Theorem

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 $u,v,y_1,y_2$

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November 2, 2026 1 min read

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