Method Of Variation Of Parameters

TECHNIQUE

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

L u(x)=f(x)

for

u(x)

, where

L

and

f(x)

are known.

Suppose we have two linearly independent solutions to the given differential equation, ''u''₁ and ''u''₂. Let ''W'' be the Wronskian of these two functions, and ''W'' must be nonzero since we have supposed the solutions are linearly independent.

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 ''u''₁ and ''u''₂ 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

: $\begin{pmatrix}
A'(x) \
B'(x)\end{pmatrix}=
\begin{pmatrix}
u_1(x) & u_2(x) \
u_1'(x) & u_2'(x) \end{pmatrix}^{-1}
\begin{pmatrix}
0\
f\end{pmatrix}$

: $={1\over W}
\begin{pmatrix}
u_2'(x) & -u_2(x) \
-u_1'(x) & u_1(x) \end{pmatrix}
\begin{pmatrix}
0\
f\end{pmatrix}.$

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 ''in''homogeneous 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 .

EXAMPLE USAGE

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 ''u''₁=''e''^-2''x'', and ''u''₂=''xe''^{-2''x''^{. The Wronskian of these two functions is

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

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

We seek functions ''A''(''x'') and ''B''(''x'') so ''A''(''x'')''u''₁+''B''(''x'')''u''₂ 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.

A NOTE ON FIRST-ORDER DIFFERENTIAL EQUATIONS

Variation of parameters may also be used for first-order differential equations, though other methods such as Integrating Factor s and Undetermined Coefficients are usually used to obtain the same results with considerably less effort.}}}}

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