## You are here

Homevariation of parameters

## Primary tabs

# variation of parameters

The method of *variation of parameters* is a way of finding a particular
solution to a nonhomogeneous linear differential equation.

Suppose that we have an $n$th order linear differential operator

$L[y]:=y^{{(n)}}+p_{1}(t)y^{{(n-1)}}+\cdots+p_{n}(t)y,$ |

and a corresponding nonhomogeneous differential equation

$L[y]=g(t).$ | (1) |

Suppose that we know a fundamental set of solutions $y_{1},y_{2},\ldots,y_{n}$ of the corresponding homogeneous differential equation $L[y_{c}]=0$. The general solution of the homogeneous equation is

$y_{c}(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)+\cdots+c_{n}y_{n}(t),$ |

where $c_{1},c_{2},\ldots,c_{n}$ are constants. The general solution to the nonhomogeneous equation $L[y]=g(t)$ is then

$y(t)=y_{c}(t)+Y(t),$ |

where $Y(t)$ is a particular solution which satisfies $L[Y]=g(t)$, and the constants $c_{1},c_{2},\ldots,c_{n}$ are chosen to satisfy the appropriate boundary conditions or initial conditions.

The key step in using variation of parameters is to suppose that the particular solution is given by

$Y(t)=u_{1}(t)y_{1}(t)+u_{2}(t)y_{2}(t)+\cdots+u_{n}(t)y_{n}(t),$ | (2) |

where $u_{1}(t),u_{2}(t),\ldots,u_{n}(t)$ are as yet to be determined functions
(hence the name *variation of parameters*). To find
these $n$ functions we need a set of $n$ independent equations.
One obvious condition is that the proposed ansatz satisfies Eq.
(1). Many possible additional conditions are possible,
we choose the ones that make further calculations easier. Consider the
following set of $n-1$ conditions

$\displaystyle y_{1}u_{1}^{{\prime}}+y_{2}u_{2}^{{\prime}}+\cdots+y_{n}u_{n}^{{% \prime}}$ | $\displaystyle=$ | $\displaystyle 0$ | ||

$\displaystyle y_{1}^{{\prime}}u_{1}^{{\prime}}+y_{2}^{{\prime}}u_{2}^{{\prime}% }+\cdots+y_{n}^{{\prime}}u_{n}^{{\prime}}$ | $\displaystyle=$ | $\displaystyle 0$ | ||

$\displaystyle\vdots$ | ||||

$\displaystyle y_{1}^{{(n-2)}}u_{1}^{{\prime}}+y_{2}^{{(n-2)}}u_{2}^{{\prime}}+% \cdots+y_{n}^{{(n-2)}}u_{n}^{{\prime}}$ | $\displaystyle=$ | $\displaystyle 0.$ |

Now, substituting Eq. (2) into $L[Y]=g(t)$ and using the above conditions, we can get another equation

$y_{1}^{{(n-1)}}u_{1}^{{\prime}}+y_{2}^{{(n-1)}}u_{2}^{{\prime}}+\cdots+y_{n}^{% {(n-1)}}u_{n}^{{\prime}}=g.$ |

So we have a system of $n$ equations for $u_{1}^{{\prime}},u_{2}^{{\prime}},\ldots,u_{n}^{{\prime}}$ which we can solve using Cramer’s rule:

$u_{m}^{{\prime}}(t)=\frac{g(t)W_{m}(t)}{W(t)},\quad m=1,2,\ldots,n.$ |

Such a solution always exists since the Wronskian $W=W(y_{1},y_{2},\ldots,y_{n})$ of the system is nowhere zero, because the $y_{1},y_{2},\ldots,y_{n}$ form a fundamental set of solutions. Lastly the term $W_{m}$ is the Wronskian determinant with the $m$th column replaced by the column $(0,0,\ldots,0,1)$.

Finally the particular solution can be written explicitly as

$Y(t)=\sum_{{m=1}}^{n}y_{m}(t)\int\frac{g(t)W_{m}(t)}{W(t)}dt.$ |

# References

- 1 W. E. Boyce and R. C. DiPrima. Elementary Differential Equations and Boundary Value Problems John Wiley & Sons, 6th edition, 1997.

## Mathematics Subject Classification

34A30*no label found*34A05

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections

## Comments

## variation of parameters

I am not sure why this item lies in the "lacking a proof" list. Perhaps it is because of the title, which implies an ad-hoc method of solution. If one approaches this through matrix calculus, having discussed the solution of dZ/dt = MZ with a functional coefficient matrix M(t) and full matrix of solutions Z(t), the way to approach the inhomogeneous equation dW/dt = MW + F is to factor the solution matrix, writing W = ZQ.

Then dW/dt = dZ/dt Q + Z dQ/dt = MZQ + F. Supposing now that the homogeneous equation has already been solved, and therefore a known quantity, it remains to solve Z dQ/dt = F. A result from matrix calculus assures us that Z is invertible if it evolves from a basis of initial conditions, leaving the quadrature (that is, pure integration, no solving) Q = Q(0) + Integral(0,t)[Z^-1(s)F(s)ds].

Since the inverse of a matrix is its adjugate divided by its determinant (which in the case of an n'th order equation is a Wronskian), there results the nice formula in this posting.

Admittedly to convert the above into a proof, some details from matrix calculus have to be assumed, such as the existence of a solution in the form of the matrizant (Picard's method) or as a product integral (Euler's method). Likewise that d|Z|/dt = Trace(Z)|Z|, by having an exponential solution which never vanishes for analytic Z, guarantees the invertibility of Z.

- hvm