21. Variation of Parameters for Higher-order Equations

Dated: 02-12-2024

Step 1

To find the complementary function, we solve the associated homogeneous equation

\[a_n\frac{d^ny}{dx^n}+a_{n-1}\frac{d^{n-1}y}{dx^{n-1}}+\cdots+a_1\frac{dy}{dx}+a_0y=0\]

Step 2

Suppose that the complementary function for the equation is

\[y=c_1y_1+c_2y_2+\ldots+c_ny_n\]

Then \(y_1, y_2, \ldots, y_n\) are \(n\) linearly independent solutions of the homogeneous equation.
Therefore, we compute Wronskian of these solutions.

\[W = (y_1, y_2, y_3, \ldots, y_n) = \begin{vmatrix} y_1 & y_2 & \cdots & y_n \\ y_1' & y_2' & \cdots & y_n' \\ \vdots & \vdots & \ddots & \vdots \\ y_1^{(n-1)} & y_2^{(n - 1)} & \ldots & y_n^{(n - 1)} \end{vmatrix} \]

Step 3

We write the differential equation in the form

\[y^{(n)}+P_{n-1}(x)y^{(n-1)}+\cdots+P_1(x)y'+P_0(x)y=f(x)\]

and compute the determinants \(W_k\), \(k = 1, 2, \ldots, n\) by replacing the \(k\) column of \(W\) by the column

\[ \begin{matrix} 0 \\ 0 \\ \vdots \\ f(x) \end{matrix} \]

Step 4

Next we find the derivatives \(u_1', u_2', \ldots, u_n'\) of the unknown functions \(u_1, u_2, \ldots, u_n\) through the relations

\[u_k' = \frac{W_k}{W}, \quad k=1,2,…,n\]

Note that these derivatives can be found by solving the \(n\) equations.

\[ \begin{matrix} y_1u_1'& + &y_2u_2'& +\cdots+&y_nu_n'&=0 \\ y_1'u_1'& + &y_2'u_2'& +\cdots+&y_n'u_n'&=0 \\ \vdots & &\vdots& &\vdots \\ y_1^{(n-1)}u_1'& + &y_2^{(n-1)}u_2'&+\cdots+&y_n^{(n-1)}u_n'& =f(x) \end{matrix} \]

Step 5

Integrate the derivative functions computed in the step 4 to find the functions \(u_k\).

\[u_k=\int\frac{W_k}{W}dx, \quad k=1,2,…,n\]

Step 6

We write a particular solution of the given non-homogeneous equation as

\[y_p=u_1(x)y_1(x)+u_2(x)y_2(x)+\cdots+u_n(x)y_n(x)\]

Step 7

Having found the complementary function \(y_c\) and the particular integral \(y_p\), we write the general solution by substitution in the expression

\[y = y_c + y_p\]

Note

The first \(n - 1\) equations in step 4 are assumptions made to simplify the first \(n - 1\) derivatives \(y_p\).
The last equation in the system results from substituting the particular integral \(y_p\) and its derivatives into the given \(nth\) order linear differential equation and then simplifying.
Depending on how the integrals of the derivatives \(u_k'\) of the unknown functions are found, the answer for \(y_p\) may be different for different attempts to find \(y_p\) for the same equation.
When asked to solve an initial value problem, we need to be sure to apply the initial conditions to the general solution and not to the complementary function alone, thinking that it is only \(y_c\) that involves the arbitrary constants.

Example

Solve the differential equation by variation of parameters.

\[y'''-2y''-y'+2y=e^{3x}\]

Solution

Step 1

The associated homogeneous equation is

\[y'''-2y''-y'+2y=0\]

The auxiliary equation of the homogeneous differential equation is

\[m^3-2m^2-m+2=0\]

\[\Rightarrow (m-2)(m^2-1)=0\]

\[\Rightarrow m=1,2,-1\]

The roots of the auxiliary equation are real and distinct.
Therefore \(y_c\) is given by

\[y_c=c_1e^{x}+c_2e^{2x}+c_3e^{-x}\]

Step 2

From \(y_c\) we find that three linearly independent solutions of the homogeneous differential equation.

\[y_1=e^x, y_2=e^{2x}, y_3=e^{-x}\]

Thus the Wronskian of the solutions \(y_1, y_2\) and \(y_3\) is given by

\[W= \begin{vmatrix} e^{x}&e^{2x}&e^{-x}\\ e^{x}&2e^{2x}&-e^{-x}\\ e^{x}&4e^{2x}&e^{-x} \end{vmatrix} = e^{x}\cdot e^{2x}\cdot e^{-x} \begin{vmatrix} 1&1&1\\ 1&2&-1\\ 1&4&1 \end{vmatrix} \]

By applying the row operations \(R_2 - R_1, R_3 - R_1\), we obtain

\[W= e^{2x} \begin{vmatrix} 1&1&1\\ 0&1&-2\\ 0&3&0 \end{vmatrix} =6e^{2x}\neq 0 \]

Step 3

The given differential equation is already in the required standard form

\[y'''-2y''-y'+2y=e^{3x}\]

Step 4

Next we find the determinants \(W_1, W_2\) and \(W_3\) by, respectively, replacing the \(1^{st}, 2^{nd}\) and \(3^{rd}\) column of \(W\) by the column

\[ \begin{matrix} 0 \\ 0 \\ e^{3x} \end{matrix} \]

Thus

\[ W_1= \begin{vmatrix} 0&e^{2x}&e^{-x}\\ 0&2e^{2x}&-e^{-x}\\ e^{3x}&4e^{2x}&e^{-x} \end{vmatrix} =(-1)^{2+1} \begin{vmatrix} e^{2x}&e^{-x}\\ 2e^{2x}&-e^{-x} \end{vmatrix} e^{3x} \]

\[=e^{3x}(-e^{x}-2e^{x})=-3e^{4x}\]

\[ W_2= \begin{vmatrix} e^{x}&0&e^{-x}\\ e^{x}&0&-e^{-x}\\ e^{x}&e^{3x}&e^{-x} \end{vmatrix} =(-1)^{3+2} \begin{vmatrix} e^{x}&e^{-x}\\ e^{x}&e^{-x} \end{vmatrix} e^{3x} \]

\[=-(-e^{0}-e^{0})e^{3x}=2e^{3x}\]

\[ W_3= \begin{vmatrix} e^{x}&e^{2x}&0\\ e^{x}&2e^{2x}&0\\ e^{x}&4e^{2x}&e^{3x} \end{vmatrix} =e^{3x} \begin{vmatrix} e^{x}&e^{2x}\\ e^{x}&2e^{2x} \end{vmatrix} \]

\[=e^{3x}(2e^{3x}-e^{3x})=e^{6x}\]

Step 5

Therefore, the derivatives of the unknown functions \(u_1, u_2\) and \(u_3\) are given by

\[u_1'=\frac{W_1}{W}=\frac{-3e^{4x}}{6e^{2x}}=-\frac{1}{2}e^{2x}\]

\[u_2'=\frac{W_2}{W}=\frac{2e^{3x}}{6e^{2x}}=\frac{1}{3}e^{x}\]

\[u_3'=\frac{W_3}{W}=\frac{e^{6x}}{6e^{2x}}=\frac{1}{6}e^{4x}\]

Step 6

Integrate these derivatives to find \(u_1, u_2\) and \(u_3\).

\[u_1=\int\frac{W_1}{W}dx=\int-\frac{1}{2}e^{2x}dx=-\frac{1}{2}\int e^{2x}dx=-\frac{1}{4}e^{2x}\]

\[u_2=\int\frac{W_2}{W}dx=\int\frac{1}{3}e^{x}dx=\frac{1}{3}e^{x}\]

\[u_3=\int\frac{W_3}{W}dx=\int\frac{1}{6}e^{4x}dx=\frac{1}{24}e^{4x}\]

Step 7

A particular solution of the non-homogeneous equation is

\[y_p=-\frac{1}{4}e^{3x}+\frac{1}{3}e^{3x}+\frac{1}{24}e^{3x}\]

Step 8

The general solution of the given differential equation is

\[y=c_1e^{x}+c_2e^{2x}+c_3e^{-x}-\frac{1}{4}e^{3x}+\frac{1}{3}e^{3x}+\frac{1}{24}e^{3x}\]

References

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