 Variation of parameters

Variation of parameters replaces the constants in a complementary solution by functions to construct a particular solution.

For the inhomogeneous linear differential equation $y^{\prime \prime }+p{y}^{\prime }+qy=r$, assume $p,q,r$ are continuous functions on an interval $I$, and $y_1,y_2$ are linearly independent homogeneous solutions there. With $W=W(y_1,y_2)$ and $x_0\in I$, variation of parameters constructs the particular solutionThe Wronskian has no zeros on $I$, so these quotients are defined. For $A{y}^{\prime \prime }+B{y}^{\prime }+Cy=F$, normalize first: $r=F/A$, not $F$.^[ref]

To derive the formula, write $y_p=u_1{y}_1+u_2{y}_2$ and impose $u_1^{\prime }{y}_1+u_2^{\prime }{y}_2=0$. This auxiliary choice removes extra terms from the first derivative. Substitution into the inhomogeneous linear differential equation leaves $u_1^{\prime }{y}_1^{\prime }+u_2^{\prime }{y}_2^{\prime }=r$. The two linear equations for $u_1^{\prime },u_2^{\prime }$ have determinant $W$, and Cramer's rule givesIntegrating from $x_0$ makes $y_p(x_0)=y_p^{\prime }(x_0)=0$. Arbitrary integration constants contribute only homogeneous solutions; adding $c_1{y}_1+c_2{y}_2$ supplies the general solution.

For a worked example, use the same homogeneous solutions $e^{-x},e^{2x}$ and the forcing term $r=6e^{3x}$ in $y^{\prime \prime }-y^{\prime }-2y=6e^{3x}$. With $x_0=0$ and $W=3e^x$,Consequently the particular solution selected by zero initial conditions isIts value and first derivative vanish at zero. Direct substitution confirms the forcing term: the differential operator $L[y]=y^{\prime \prime }-y^{\prime }-2y$ annihilates the last two terms and maps $e^{3x}$ to $4e^{3x}$. The two integrals therefore build a response from known homogeneous solutions, even when the forcing term has no convenient elementary guess.