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7.4: Variation of Parameters for Higher Order Equations

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Nov 27, 2022
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- 7.3.1: Annihilation for Higher Order Equations (Exercises)
- 7.4.1: Variation of Parameters for Higher Order Equations (Exercises)

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Derivation of the Method

Caution

If you have not had Math 410 (linear algebra), then you may want to review Appendix 11.4 again before starting this section.

Although the method of variation of parameters for higher order equations is similar to what we did in Section 5.6 for second order equations, there are differences that we'll need to discuss here. For convenience, we consider linear differential equations written as

$P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=F(x),$

which can be rewritten as

$\label{eq:9.4.1} y^{(n)}+p_1(x)y^{(n-1)}+\cdots+p_n(x)y=f(x).$

on any interval on which $P_0$ has no zeros, with $p_1=P_1/P_0$ , …, $p_n=P_n/P_0$ and $f=F/P_0$ . For simplicity, throughout this chapter we’ll abbreviate the left side of Equation $\ref{eq:9.4.1}$ by $Ly$ ; that is,

$Ly=y^{(n)}+p_1(x)y^{(n-1)}+\cdots+p_n(x)y.\nonumber$

When we speak of solutions of this equation and its homogeneous equation $Ly=0$ , we mean solutions on $(a,b)$ . We’ll show how to use the method of variation of parameters to find a particular solution of $Ly=f(x)$ , provided that we know a fundamental set of solutions $\{y_1,y_2,\dots,y_n\}$ of $Ly=0$ .

We seek a particular solution of $Ly=f(x)$ in the form

$\label{eq:9.4.2} y_p=u_1y_1+u_2y_2+\cdots+u_ny_n$

where $\{y_1,y_2,\dots,y_n\}$ is a known fundamental set of solutions of the homogeneous equation

$y^{(n)}+p_1(x)y^{(n-1)}+\cdots+p_n(x)y=0\nonumber$

and $u_1$ , $u_2$ , …, $u_n$ are functions to be determined. We begin, as we did in Section 5.6, by imposing conditions that make our calculations easier. These conditions are similar to what we did in Section 5.6.

We impose the following $n-1$ conditions on $u_1,u_2,\dots,u_n$ :

$\label{eq:9.4.3} \begin{array}{rcl} u'_1y_1+u'_2y_2+&\cdots&+u'_ny_n=0 \\ u'_1y'_1+u'_2y'_2+&\cdots&+u'_ny'_n=0 \\ \phantom{u'_1y^{(n_1)}+u'_2y_2^{(n-1)}}&\vdots& \phantom{\cdots+u'_ny^{(n-1)}_n=q} \\ u'_1y_1^{(n-2)}+u'_2y^{(n-2)}_2+&\cdots&+u'_ny^{(n-2)}_n =0. \end{array}$

These conditions lead to simple formulas for the first $n-1$ derivatives of $y_p$ :

$\label{eq:9.4.4} y^{(r)}_p=u_1y^{(r)}_1+u_2y_2^{(r)}\cdots+u_ny^{(r)}_n,\ 0 \le r \le n-1.$

These formulas are easy to remember, since they look as though we obtained them by differentiating Equation $\ref{eq:9.4.2}$ $n-1$ times while treating $u_1$ , $u_2$ , …, $u_n$ as constants. To see that Equation $\ref{eq:9.4.3}$ implies Equation $\ref{eq:9.4.4}$ , we first differentiate Equation $\ref{eq:9.4.2}$ to obtain

$y_p'=u_1y_1'+u_2y_2'+\cdots+u_ny_n'+u_1'y_1+u_2'y_2+\cdots+u_n'y_n,\nonumber$

which reduces to

$y_p'=u_1y_1'+u_2y_2'+\cdots+u_ny_n'\nonumber$

because of the first equation in Equation $\ref{eq:9.4.3}$ . Differentiating this yields

$y_p''=u_1y_1''+u_2y_2''+\cdots+u_ny_n''+u_1'y_1'+u_2'y_2'+\cdots+u_n'y_n',\nonumber$

which reduces to

$y_p''=u_1y_1''+u_2y_2''+\cdots+u_ny_n''\nonumber$

because of the second equation in Equation $\ref{eq:9.4.3}$ . Continuing in this way yields Equation $\ref{eq:9.4.4}$ .

The last equation in Equation $\ref{eq:9.4.4}$ is

$y_p^{(n-1)}=u_1y_1^{(n-1)}+u_2y_2^{(n-1)}+\cdots+u_ny_n^{(n-1)}.\nonumber$

Differentiating this yields

$y_p^{(n)}=u_1y_1^{(n)}+u_2y_2^{(n)}+\cdots+u_ny_n^{(n)}+ u_1'y_1^{(n-1)}+u_2'y_2^{(n-1)}+\cdots+u_n'y_n^{(n-1)}.\nonumber$

Substituting this and Equation $\ref{eq:9.4.4}$ into Equation $\ref{eq:9.4.1}$ yields

$u_1Ly_1+u_2Ly_2+\cdots+u_nLy_n+\left( u_1'y_1^{(n-1)}+u_2'y_2^{(n-1)}+\cdots+u_n'y_n^{(n-1)}\right)=f(x).\nonumber$

Since $Ly_i=0$ $(1 \le i \le n)$ , this reduces to

$u_1'y_1^{(n-1)}+u_2'y_2^{(n-1)}+\cdots+u_n'y_n^{(n-1)}=f(x).\nonumber$

Combining this equation with Equation $\ref{eq:9.4.3}$ shows that

$y_p=u_1y_1+u_2y_2+\cdots+u_ny_n\nonumber$

is a solution of Equation $\ref{eq:9.4.1}$ if

$\begin{array}{rcl} u'_1y_1+u'_2y_2+&\cdots&+u'_ny_n=0 \\ u'_1y'_1+u'_2y'_2+&\cdots&+u'_ny'_n=0 \\ \phantom{u'_1y^{(n_1)}+u'_2y_2^{(n-1)}}&\vdots& \phantom{\cdots+u'_ny^{(n-1)}_n=q} \\ u'_1y_1^{(n-2)}+u'_2y^{(n-2)}_2+&\cdots&+u'_ny^{(n-2)}_n =0 \\ u'_1y^{(n-1)}_1+u'_2y^{(n-1)}_2+&\cdots&+u'_n y^{(n-1)}_n=f(x), \end{array}\nonumber$

where $W=\label{eq:9.4.5} \left[\begin{array}{cccc} y_1&y_2&\cdots&y_n \\[4pt] y'_1&y'_2&\cdots&y_n'\\[4pt] \vdots&\vdots&\ddots&\vdots\\[4pt] y_1^{(n-2)}&y_2^{(n-2)}&\cdots&y_n^{(n-2)}\\[4pt] y_1^{(n-1)}&y_2^{(n-1)}&\cdots&y_n^{(n-1)} \end{array} \right]$ and b= $\left[\begin{array}{c}0\\0\\ \vdots\\0\\f(x)\end{array}\right].$

The determinant of $W$ is the Wronskian of the fundamental set of solutions $\{y_1,y_2,\dots,y_n\}$ , which we know has no zeros on $(a,b)$ since the solutions are linearly independent. Solving by Cramer’s rule yields

$\label{eq:9.4.6} u'_j={W_j\over W},\quad 1\le j\le n,$

where $W_j$ is the determinant obtained by replacing the $j$ -th column of $W$ with b.

Having obtained $u_1'$ , $u_2'$ $, \dots,$ $u_n'$ , we can integrate to obtain $u_1,\,u_2,\dots,u_n$ . As in Section 5.6, we take the constants of integration to be zero, and we drop any linear combination of $\{y_1,y_2,\dots,y_n\}$ that may appear in $y_p$ .

Third Order Equations

If $n=3$ , then

$W=\left| \begin{array}{ccc} y_1&y_2&y_3 \\[4pt] y'_1&y'_2&y'_3 \\[4pt] y''_1&y''_2&y''_3 \end{array} \right|.\nonumber$

Therefore

$W_1=\left| \begin{array}{cc} 0&y_2&y_3 \\[4pt] 0&y'_2&y'_3 \\[4pt] f(x)&y_2''&y_3''\end{array} \right|, \quad W_2=\left| \begin{array}{cc} y_1&0&y_3 \\[4pt] y'_1&0&y'_3 \\[4pt] y_1''&f(x)&y_3'' \end{array} \right|, \quad W_3=\left| \begin{array}{cc} y_1&y_2&0 \\[4pt] y'_1&y'_2&0 \\[4pt] y_1''&y_2''&f(x) \end{array} \right|,\nonumber$

and Equation $\ref{eq:9.4.6}$ becomes

$\label{eq:9.4.7} u'_1={W_1\over W},\quad u'_2={W_2\over W},\quad u'_3={W_3\over W}.$

Example 7.4.1

Find the general solution of

$\label{eq:9.4.8} xy'''-y''-xy'+y=8x^2e^x,$

given that $y_1=x$ , $y_2=e^x$ , and $y_3=e^{-x}$ form a fundamental set of solutions of the homogeneous equation. Then find the general solution of Equation $\ref{eq:9.4.8}$ .

Solution

We seek a particular solution of Equation $\ref{eq:9.4.8}$ of the form

$y_p=u_1x+u_2e^x+u_3e^{-x}.\nonumber$

The Wronskian of $\{y_1,y_2,y_3\}$ is

$W(x)=\left| \begin{array}{ccr} x&e^x&e^{-x} \\ 1&e^x&-e^{-x} \\ 0&e^x&e^{-x} \end{array} \right|=2x,\nonumber$

$\begin{aligned} W_1&= \left| \begin{array}{cr} 0&e^x&e^{-x}\\ 0&e^x&-e^{-x}\\8xe^x&e^x&e^{-x} \end{array} \right|=-16xe^x,\quad W_2&= \left| \begin{array}{cr} x&0&e^{-x}\\ 1&0&-e^{-x}\\0&8xe^x&e^{-x} \end{array} \right|=8x(x+1),\quad W_3&= \left| \begin{array}{cc} x&e^x&0\\ 1&e^x&0\\0&e^x&8xe^x \end{array} \right|=8xe^{2x}(x-1).\end{aligned}$

Therefore, from Equation $\ref{eq:9.4.7}$

$\begin{aligned} u'_1={W_1\over W}=-8e^x,\quad u_2'={W_2\over W}=4(x+1),\quad u_3'={W_3\over W}=4e^{2x}(x-1).\end{aligned}\nonumber$

Integrating and taking the constants of integration to be zero yields

$u_1=-8e^x,\quad u_2=2(x+1)^2, u_3=e^{2x}(2x-3).\nonumber$

Hence,

$\begin{aligned} y_p&=u_1y_1+u_2y_2+u_3y_3\\ &=(-8e^x)x+e^x(2(x+1)^2)+e^{-x}\left(e^{2x}(2x-3)\right) \\&=e^x(2x^2-2x-1).\end{aligned}$

Since $-e^x$ is a solution of the complementary equation, we redefine

$y_p=2xe^x(x-1).\nonumber$

Therefore the general solution of Equation $\ref{eq:9.4.8}$ is

$y=2xe^x(x-1)+c_1x+c_2e^x+c_3e^{-x}.\nonumber$

Fourth Order Equations

If $n=4$ , then

$W=\left| \begin{array}{cccc} y_1&y_2&y_3&y_4 \\[4pt] y'_1&y'_2&y'_3&y_4' \\[4pt] y''_1&y''_2&y''_3&y_4''\\[4pt] y'''_1&y'''_2&y'''_3&y_4''' \end{array} \right|,\nonumber$

Therefore

$W_1=\left| \begin{array}{cccc} 0&y_2&y_3&y_4 \\[4pt] 0&y'_2&y'_3&y_4'\\[4pt] 0&y''_2&y''_3&y_4''\\[4pt] f(x)&y_2'''&y_3'''&y_4''' \end{array} \right|,\quad W_2=\left| \begin{array}{cccc} y_1&0&y_3&y_4 \\[4pt] y'_1&0&y'_3&y_4'\\[4pt] y''_1&0&y''_3&y_4''\\[4pt]y_1'''&f(x)& y_3'''&y_4'''\end{array} \right|,\nonumber$

$W_3=\left| \begin{array}{cccc} y_1&y_2&0&y_4 \\[4pt] y'_1&y'_2&0&y_4'\\[4pt] y''_1&y''_2&0&y_4''\\[4pt]y_1'''&y_2'''&f(x)&y_4''' \end{array} \right|,\quad W_4=\left| \begin{array}{cccc} y_1&y_2&y_3&0 \\[4pt] y_1'&y'_2&y_3'&0\\[4pt] y_1''&y''_2&y_3''&0\\[4pt]y_1'''&y_2'''&y_3'''&f(x) \end{array} \right|,\nonumber$

and Equation $\ref{eq:9.4.6}$ becomes

$\label{eq:9.4.9} u'_1={W_1\over W},\quad u'_2={W_2\over W},\quad u'_3={W_3\over W},\quad u'_4={W_4\over W}.$

Example 7.4.2

Find the general solution of

$\label{eq:9.4.10} x^4y^{(4)}+6x^3y'''+2x^2y''-4xy'+4y=12x^2,$

given that $y_1=x$ , $y_2=x^2$ , $y_3=1/x$ and $y_4=1/x^2$ form a fundamental set of solutions of the homogeneous equation. Then find the general solution of Equation $\ref{eq:9.4.10}$ on $(-\infty,0)$ and $(0,\infty)$ .

Solution

We seek a particular solution of Equation $\ref{eq:9.4.10}$ of the form

$y_p=u_1x+u_2x^2+{u_3\over x}+{u_4\over x^2}.\nonumber$

The Wronskian of $\{y_1,y_2,y_3,y_4\}$ is

$W(x)=\left| \begin{array}{cccr} x&x^2&1/x&-1/x^2 \\[4pt] 1&2x&-1/x^2&-2/x^3 \\[4pt] 0 &2&2/x^3&6/x^4\\[4pt] 0&0&-6/x^4&-24/x^5 \end{array} \right|=-72/x^6,\nonumber$

$W_1=\left| \begin{array}{cccc} 0&x^2&1/x&1/x^2\\[4pt]0&2x&-1/x^2&-2/x^3\\[4pt] 0&2&2/x^3&6/x^4\\[4pt]12/x^2&0&-6/x^4&-24/x^5, \end{array} \right|=144/x^6,\quad W_2= \left| \begin{array}{cccc} x&0&1/x&1/x^2\\[4pt]1&0&-1/x^2&-2/x^3\\[4pt] 0&0&2/x^3&6/x^4\\[4pt] 0&12/x^2&-6/x^4&-24/x^5\end{array} \right|=-72/x^7,\nonumber$

$W_3= \left| \begin{array}{cccc} x&x^2&0&-1/x^2 \\[4pt] 1&2x&0&-2/x^3 \\[4pt] 0 &2&0&6/x^4\\[4pt] 0&0&12/x^2&-24/x^5 \end{array} \right|=-144/x^4,\quad W_4= \left| \begin{array}{ccc} x&x^2&1/x&0 \\[4pt] 1&2x&-1/x^2&0 \\[4pt] 0 &2&2/x^3&0\\[4pt] 0&0&-6/x^4&12/x^2 \end{array} \right|=72/x^3,\nonumber$

Therefore, from Equation $\ref{eq:9.4.9}$ ,

$\begin{aligned} u'_1= {W_1\over W}=-2,\quad u_2'={W_2\over W}=1/x,\quad u_3'={W_3\over W}=2x^2,\quad u_4'={W_4\over W}=-x^3.\end{aligned}$

Integrating these and taking the constants of integration to be zero yields

$u_1=-2x,\quad u_2=\ln|x|,\quad u_3={2x^3\over3}, u_4=-{x^4\over4}.\nonumber$

Hence,

$\begin{aligned} y_p&=u_1y_1+u_2y_2+u_3y_3+u_4y_4\\ &=(-2x)x+(\ln|x|)x^2+{2x^3\over3}{1\over x}+\left(-{x^4\over4}\right) {1\over x^2} \\&=x^2\ln|x|-{19x^2\over12}.\end{aligned}$

Since $-19x^2/12$ is a solution of the homogeneous equation, we redefine

$y_p=x^2\ln|x|.\nonumber$

Therefore

$y=x^2\ln|x|+c_1x+c_2x^2+{c_3\over x}+{c_4\over x^2}\nonumber$

is the general solution of Equation $\ref{eq:9.4.10}$ on $(-\infty,0)$ and $(0,\infty)$ .

Derivation of the Method

Caution

Third Order Equations

Solution

Fourth Order Equations

Solution

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