Variation of parameters

In mathematics, variation of parameters also known as variation of constants, is a general method to solve inhomogeneous linear ordinary differential equations. It was developed by the Italian-French mathematician Joseph Louis Lagrange.

For first-order inhomogeneous linear differential equations it's usually possible to find solutions via integrating factors or undetermined coefficients with considerably less effort, although those methods are rather heuristics that involve guessing and don't work for all inhomogenous linear differential equations.

Given an ordinary non-homogeneous linear differential equation of order n

$y^{(n)}(x)+\sum _{i=0}^{n-1}a_{i}(x)y^{(i)}(x)=b(x)$. (i)

let $y_{1}(x),\ldots ,y_{n}(x)$ be a fundamental system of the corresponding homogeneous equation

$y^{(n)}(x)+\sum _{i=0}^{n-1}a_{i}(x)y^{(i)}(x)=0.$ (ii)

Then a particular solution to the non-homogeneous equation is given by

$\sum _{i=1}^{n}c_{i}(x)y_{i}(x)$ (iii)

where the $c_{i}(x)$ are continuous functions which satisfy the equations

$\sum _{i=1}^{n}c_{i}^{'}(x)y_{i}^{(j)}(x)=0\,\mathrm {,} \quad j=0,\ldots ,n-2$ (iv)

(results from substitution of (iii) into the homogeneous case (ii); )

and

$\sum _{i=1}^{n}c_{i}^{'}(x)y_{i}^{(n-1)}(x)=b(x).$. (v)

(results from substitution of (iii) into (i) and applying (iv);
$c_{i}'(x)=0$ for all x and i is the only way to satisfy the condition, since all $y_{i}(x)$ are linearly independent. It implies that all $c_{i}(x)$ are independent of x in the homogeneous case b(x)=0. )

This linear system of n equations can then be solved using Cramer's rule yielding

$c_{i}^{'}(x)={\frac {W_{i}(x)}{W(x)}}\,\mathrm {,} \quad i=1,\ldots ,n$

where $W(x)$ is the Wronskian determinant of the fundamental system and $W_{i}(x)$ is the Wronskian determinant of the fundamental system with the i-th column replaced by $(0,0,\ldots ,b(x)).$

The particular solution to the non-homogeneous equation can then be written as

$\sum _{i=1}^{n}\int {\frac {W_{i}(x)}{W(x)}}dx\,y_{i}(x)$.

Examples

Specific second order equation

Let us solve

$y''+4y'+4y=\cosh {x}.\;\!$

We want to find the general solution to the differential equation, that is, we want to find solutions to the homogeneous differential equation

$y''+4y'+4y=0.\;\!$

Form the characteristic equation

$\lambda ^{2}+4\lambda +4=(\lambda +2)^{2}=0\;\!$

$\lambda =-2,-2.\;\!$

Since we have a repeated root, we have to introduce a factor of x for one solution to ensure linear independence.

So, we obtain u₁=e^-2x, and u₂=xe^-2x. The Wronskian of these two functions is

${\begin{vmatrix}e^{-2x}&xe^{-2x}\\-2e^{-2x}&-e^{-2x}(2x-1)\\\end{vmatrix}}=-e^{-2x}e^{-2x}(2x-1)+2xe^{-2x}e^{-2x}$

$=-e^{-4x}(2x-1)+2xe^{-4x}=(-2x+1+2x)e^{-4x}=e^{-4x}.\;\!$

Because the Wronskian is non-zero, the general solution found for the differential equation is linearly independent so it is correct.

We seek functions A(x) and B(x) so A(x)u₁+B(x)u₂ is a general solution of the inhomogeneous equation. We need only calculate the integrals

$A(x)=-\int {1 \over W}u_{2}(x)f(x)\,dx,\;B(x)=\int {1 \over W}u_{1}(x)f(x)\,dx$

that is,

$A(x)=-\int {1 \over e^{-4x}}xe^{-2x}\cosh {x}\,dx=-\int xe^{2x}\cosh {x}\,dx=-{1 \over 18}e^{x}(9(x-1)+e^{2x}(3x-1))+C_{1}$

$B(x)=\int {1 \over e^{-4x}}e^{-2x}\cosh {x}\,dx=\int e^{2x}\cosh {x}\,dx={1 \over 6}e^{x}(3+e^{2x})+C_{2}$

where $C_{1}$ and $C_{2}$ are constants of integration.

General second order equation

We have a differential equation of the form

$u''+p(x)u'+q(x)u=f(x)\,$

and we define the linear operator

$L=D^{2}+p(x)D+q(x)\,$

where D represents the differential operator. We therefore have to solve the equation $Lu(x)=f(x)$ for $u(x)$, where $L$ and $f(x)$ are known.

We must solve first the corresponding homogeneous equation:

$u''+p(x)u'+q(x)u=0\,$

by the technique of our choice. Once we've obtained two linearly independent solutions to this homogeneous differential equation (because this ODE is second-order) — call them u₁ and u₂ — we can proceed with variation of parameters.

Now, we seek the general solution to the differential equation $u_{G}(x)$ which we assume to be of the form

$u_{G}(x)=A(x)u_{1}(x)+B(x)u_{2}(x).\,$

Here, $A(x)$ and $B(x)$ are unknown and $u_{1}(x)$ and $u_{2}(x)$ are the solutions to the homogeneous equation. Observe that if $A(x)$ and $B(x)$ are constants, then $Lu_{G}(x)=0$. We desire A=A(x) and B=B(x) to be of the form

$A'(x)u_{1}(x)+B'(x)u_{2}(x)=0.\,$

Now,

$u_{G}'(x)=(A(x)u_{1}(x)+B(x)u_{2}(x))'=(A(x)u_{1}(x))'+(B(x)u_{2}(x))'\,$

$=A'(x)u_{1}(x)+A(x)u_{1}'(x)+B'(x)u_{2}(x)+B(x)u_{2}'(x)\,$

$=A'(x)u_{1}(x)+B'(x)u_{2}(x)+A(x)u_{1}'(x)+B(x)u_{2}'(x)\,$

and since we have required the above condition, then we have

$u_{G}'(x)=A(x)u_{1}'(x)+B(x)u_{2}'(x).\,$

Differentiating again (omitting intermediary steps)

$u_{G}''(x)=A(x)u_{1}''(x)+B(x)u_{2}''(x)+A'(x)u_{1}'(x)+B'(x)u_{2}'(x).\,$

Now we can write the action of L upon u_G as

$Lu_{G}=A(x)Lu_{1}(x)+B(x)Lu_{2}(x)+A'(x)u_{1}'(x)+B'(x)u_{2}'(x).\,$

Since u₁ and u₂ are solutions, then

$Lu_{G}=A'(x)u_{1}'(x)+B'(x)u_{2}'(x).\,$

We have the system of equations

${\begin{pmatrix}u_{1}(x)&u_{2}(x)\\u_{1}'(x)&u_{2}'(x)\end{pmatrix}}{\begin{pmatrix}A'(x)\\B'(x)\end{pmatrix}}={\begin{pmatrix}0\\f\end{pmatrix}}.$

Expanding,

${\begin{pmatrix}A'(x)u_{1}(x)+B'(x)u_{2}(x)\\A'(x)u_{1}'(x)+B'(x)u_{2}'(x)\end{pmatrix}}={\begin{pmatrix}0\\f\end{pmatrix}}.$

So the above system determines precisely the conditions

$A'(x)u_{1}(x)+B'(x)u_{2}(x)=0\,$

$A'(x)u_{1}'(x)+B'(x)u_{2}'(x)=Lu_{G}=f.\,$

We seek A(x) and B(x) from these conditions, so, given

${\begin{pmatrix}u_{1}(x)&u_{2}(x)\\u_{1}'(x)&u_{2}'(x)\end{pmatrix}}{\begin{pmatrix}A'(x)\\B'(x)\end{pmatrix}}={\begin{pmatrix}0\\f\end{pmatrix}}$

we can solve for (A′(x), B′(x))^T, so

${\begin{pmatrix}A'(x)\\B'(x)\end{pmatrix}}={\begin{pmatrix}u_{1}(x)&u_{2}(x)\\u_{1}'(x)&u_{2}'(x)\end{pmatrix}}^{-1}{\begin{pmatrix}0\\f\end{pmatrix}}$

$={1 \over W}{\begin{pmatrix}u_{2}'(x)&-u_{2}(x)\\-u_{1}'(x)&u_{1}(x)\end{pmatrix}}{\begin{pmatrix}0\\f\end{pmatrix}},$

where W denotes the Wronskian of u₁ and u₂. (We know that W is nonzero, from the assumption that u₁ and u₂ are linearly independent.)

So,

$A'(x)=-{1 \over W}u_{2}(x)f(x),\;B'(x)={1 \over W}u_{1}(x)f(x)$

$A(x)=-\int {1 \over W}u_{2}(x)f(x)\,dx,\;B(x)=\int {1 \over W}u_{1}(x)f(x)\,dx.$

Whilst homogeneous equations are relatively easy to solve, this method allows the calculation of the coefficients of the general solution of the inhomogeneous equation, and thus the complete general solution of the inhomogeneous equation can be determined.

Note that $A(x)$ and $B(x)$ are each determined only up to an arbitrary additive constant (the constant of integration); one would expect two constants of integration because the original equation was second order. Adding a constant to $A(x)$ or $B(x)$ does not change the value of $Lu_{G}(x)$ because $L$ is linear.