4.2: Introduction to Linear Higher Order Equations
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)An \(n\)th order differential equation is said to be linear if it can be written in the form
\[\label{eq:9.1.1} y^{(n)}+p_1(x)y^{(n-1)}+\cdots+p_n(x)y=f(x).\]
We considered equations of this form with \(n=1\) in Section 2.1. In this section, \(n\) is an arbitrary positive integer and we sketch the general theory of linear \(n\)th order equations. We will provide some proofs for \(n=2\) in later sections.
For convenience, we consider linear differential equations written as
\[\label{eq:9.1.2} P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=F(x),\]
which can be rewritten as Equation \ref{eq:9.1.1} 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:9.1.2} by \(Ly\); that is,
\[Ly=P_0y^{(n)}+P_1y^{(n-1)}+\cdots+P_ny.\nonumber \]
We say that the equation \(Ly=F\) is normal on \((a,b)\) if \(P_0\), \(P_1\), …, \(P_n\) and \(F\) are continuous on \((a,b)\) and \(P_0\) has no zeros on \((a,b)\). If this is so then \(Ly=F\) can be written as Equation \ref{eq:9.1.1} with \(p_1\), …, \(p_n\) and \(f\) continuous on \((a,b)\).
The next theorem is analogous to Theorem 4.1.1.
Suppose \(Ly=F\) is normal on \((a,b)\), let \(x_0\) be a point in \((a,b),\) and let \(k_0\), \(k_1\), …, \(k_{n-1}\) be arbitrary real numbers\(.\) Then the initial value problem
\[Ly=F, \quad y(x_0)=k_0,\quad y'(x_0)=k_1,\dots,\quad y^{(n-1)}(x_0)=k_{n-1}\nonumber \]
has a unique solution on \((a,b)\).
Homogeneous Equations
Equation \ref{eq:9.1.2} is said to be homogeneous if \(F\equiv0\) and nonhomogeneous otherwise. Since \(y\equiv0\) is obviously a solution of \(Ly=0\), we call it the trivial solution. Any other solution is nontrivial.
If \(y_1\), \(y_2\), …, \(y_n\) are defined on \((a,b)\) and \(c_1\), \(c_2\), …, \(c_n\) are constants, then
\[\label{eq:9.1.3} y=c_1y_1+c_2y_2+\cdots+c_ny_n\]
is a linear combination of \(\{y_1,y_2\dots,y_n\}\). It’s easy to show that if \(y_1\), \(y_2\), …, \(y_n\) are solutions of \(Ly=0\) on \((a,b)\), then so is any linear combination of \(\{y_1,y_2,\dots,y_n\}\). (See the proof of Theorem 5.1.2.) We say that \(\{y_1,y_2,\dots,y_n\}\) is a fundamental set of solutions of \(Ly=0\) on \((a,b)\) if every solution of \(Ly=0\) on \((a,b)\) can be written as a linear combination of \(\{y_1,y_2,\dots,y_n\}\), as in Equation \ref{eq:9.1.3}. In this case we say that Equation \ref{eq:9.1.3} is the general solution of \(Ly=0\) on \((a,b)\).
It can be shown (Exercises 9.1.14 and 9.1.15) that if the equation \(Ly=0\) is normal on \((a,b)\) then it has infinitely many fundamental sets of solutions on \((a,b)\). The next definition will help to identify fundamental sets of solutions of \(Ly=0\).
We say that \(\{y_1,y_2,\dots,y_n\}\) is linearly independent on \((a,b)\) if the only constants \(c_1\), \(c_2\), …, \(c_n\) such that
\[\label{eq:9.1.4} c_1y_1(x)+c_2y_2(x)+\cdots+c_ny_n(x)=0,\quad a<x<b,\]
are \(c_1=c_2=\cdots=c_n=0\). If Equation \ref{eq:9.1.4} holds for some set of constants \(c_1\), \(c_2\), …, \(c_n\) that are not all zero, then \(\{y_1,y_2,\dots,y_n\}\) is linearly dependent on \((a,b)\)
The next theorem is analogous to Theorem 4.1.3.
If \(Ly=0\) is normal on \((a,b)\), then a set \(\{y_1,y_2,\dots,y_n\}\) of \(n\) solutions of \(Ly=0\) on \((a,b)\) is a fundamental set if and only if it is linearly independent on \((a,b)\).
The equation
\[\label{eq:9.1.5} x^3y'''-x^2y''-2xy'+6y=0\]
is normal and has the solutions \(y_1=x^2\), \(y_2=x^3\), and \(y_3=1/x\) on \((-\infty,0)\) and \((0,\infty)\). Show that \(\{y_1,y_2,y_3\}\) is linearly independent on \((-\infty, 0)\) and \((0,\infty)\). Then find the general solution of Equation \ref{eq:9.1.5} on \((-\infty, 0)\) and \((0,\infty)\).
Solution
Suppose
\[\label{eq:9.1.6} c_1x^2+c_2x^3+{c_3\over x}=0\]
on \((0,\infty)\). We must show that \(c_1=c_2=c_3=0\). Differentiating Equation \ref{eq:9.1.6} twice yields the system
\[\label{eq:9.1.7} \begin{array}{rr} {c_1x^2+c_2x^3+ \dfrac{c_3}{x}} &{= 0} \\ {2c_1x+3c_2x^2- \dfrac{c_3}{x^2}} &{=0} \\ {2c_1+6c_2x + \dfrac{2c_3}{x^3}} &{=0.} \end{array}\]
If Equation \ref{eq:9.1.7} holds for all \(x\) in \((0,\infty)\), then it certainly holds at \(x=1\); therefore,
\[\label{eq:9.1.8} \begin{array}{rr} {\phantom{2}c_1+\phantom{3}c_2+\phantom{2}c_3} &{= 0} \\ {2c_1+3c_2-\phantom{2}c_3}&{= 0}\\ {2c_1+6c_2+2c_3 }&{=0. }\end{array}\]
By solving this system directly, you can verify that it has only the trivial solution \(c_1=c_2=c_3=0\); however, for our purposes it is more useful to recall from linear algebra that a homogeneous linear system of \(n\) equations in \(n\) unknowns has only the trivial solution if its determinant is nonzero. Since the determinant of Equation \ref{eq:9.1.8} is
\[\left|\begin{array}{rrr}1&1&1\\2&3&-1\\2&6&2\end{array}\right|= \left|\begin{array}{rrr}1&0&0\\2&1&-3\\2&4&0\end{array}\right|=12, \nonumber\]
it follows that Equation \ref{eq:9.1.8} has only the trivial solution, so \(\{y_1,y_2,y_3\}\) is linearly independent on \((0,\infty)\). Now Theorem 4.2.2 implies that
\[y=c_1x^2+c_2x^3+{c_3\over x} \nonumber\]
is the general solution of Equation \ref{eq:9.1.5} on \((0,\infty)\). To see that this is also true on \((-\infty,0)\), assume that Equation \ref{eq:9.1.6} holds on \((-\infty,0)\). Setting \(x=-1\) in Equation \ref{eq:9.1.7} yields
\[\begin{aligned} \phantom{-2}c_1-\phantom{3}c_2-\phantom{2}c_3&=0\\ -2c_1+3c_2-\phantom{2}c_3&=0\\ \phantom{-}2c_1-6c_2-2c_3&=0.\end{aligned}\nonumber \]
Since the determinant of this system is
\[\left|\begin{array}{rrr}1&-1&-1\\-2&3&-1\\2&-6&-2\end{array}\right|= \left|\begin{array}{rrr}1&0&0\\-2&1&-3\\2&-4&0\end{array}\right|=-12, \nonumber\]
it follows that \(c_1=c_2=c_3=0\); that is, \(\{y_1,y_2,y_3\}\) is linearly independent on \((-\infty,0)\).
The equation
\[\label{eq:9.1.9} y^{(4)}+y'''-7y''-y'+6y=0\]
is normal and has the solutions \(y_1=e^x\), \(y_2=e^{-x}\), \(y_3=e^{2x}\), and \(y_4=e^{-3x}\) on \((-\infty,\infty)\). (Verify.) Show that \(\{y_1,y_2,y_3,y_4\}\) is linearly independent on \((-\infty,\infty)\). Then find the general solution of Equation \ref{eq:9.1.9}.
Solution
Suppose \(c_1\), \(c_2\), \(c_3\), and \(c_4\) are constants such that
\[\label{eq:9.1.10} c_1e^x+c_2e^{-x}+c_3e^{2x}+c_4e^{-3x}=0\]
for all \(x\). We must show that \(c_1=c_2=c_3=c_4=0\). Differentiating Equation \ref{eq:9.1.10} three times yields the system
\[\label{eq:9.1.11} \begin{array}{rcl} c_1e^x+c_2e^{-x}+\phantom{2}c_3e^{2x}+\phantom{27}c_4e^{-3x}&=&0\\ c_1e^x-c_2e^{-x}+2c_3e^{2x}-\phantom{2}3c_4e^{-3x}&=&0\\ c_1e^x+c_2e^{-x}+4c_3e^{2x}+\phantom{2}9c_4e^{-3x}&=&0\\ c_1e^x-c_2e^{-x}+8c_3e^{2x}-27c_4e^{-3x}&=&0. \end{array}\]
If Equation \ref{eq:9.1.11} holds for all \(x\), then it certainly holds for \(x=0\). Therefore
\[\begin{array}{rcl} c_1+c_2+\phantom{2}c_3+\phantom{27}c_4&=&0\\ c_1-c_2+2c_3-\phantom{2}3c_4&=&0\\ c_1+c_2+4c_3+\phantom{2}9c_4&=&0\\ c_1-c_2+8c_3-27c_4&=&0. \end{array}\nonumber \]
The determinant of this system is
\[\label{eq:9.1.12} \begin{array}{rcl} \left|\begin{array}{rrrr}1&1&1&1\\1&-1&2&-3\\1&1&4&9\\ 1&-1&8&-27\end{array}\right|&=& \left|\begin{array}{rrrr}1&1&1&1\\0&-2&1&-4\\0&0&3&8\\ 0&-2&7&-28\end{array}\right| =\left|\begin{array}{rrr}-2&1&-4\\0&3&8\\ -2&7&-28\end{array}\right|\\[5 pt]&=& \left|\begin{array}{rrr}-2&1&-4\\0&3&8 \\0&6&-24\end{array}\right|= -2\left|\begin{array}{rr}3&8\\6&-24\end{array}\right|=240, \end{array}\]
so the system has only the trivial solution \(c_1=c_2=c_3=c_4=0\). Now Theorem 4.2.2 implies that
\[y=c_1e^x+c_2e^{-x}+c_3e^{2x}+c_4e^{-3x} \nonumber\]
is the general solution of Equation \ref{eq:9.1.9}.
The Wronskian
We can use the method used in Examples 4.2.1 and 4.2.2 to test \(n\) solutions \(\{y_1,y_2,\dots,y_n\}\) of any \(n\)th order equation \(Ly=0\) for linear independence on an interval \((a,b)\) on which the equation is normal. Thus, if \(c_1\), \(c_2\),…, \(c_n\) are constants such that
\[c_1y_1+c_2y_2+\cdots+c_ny_n=0,\quad a<b,\nonumber\]
then differentiating \(n-1\) times leads to the \(n\times n\) system of equations
\[\label{eq:9.1.13} \begin{array}{rcl} c_1y_1(x)+c_2y_2(x)+&\cdots&+c_ny_n(x)=0\\ c_1y_1'(x)+c_2y_2'(x)+&\cdots&+c_ny_n'(x)=0\\ \phantom{c_1y_1^{(n)}(x)+c_2y_2^{(n-1)}(x)}&\vdots& \phantom{\cdots+c_ny_n^{(n-1)(x)}=q}\\ c_1y_1^{(n-1)}(x)+c_2y_2^{(n-1)}(x)+&\cdots&+c_ny_n^{(n-1)}(x) =0 \end{array}\]
for \(c_1\), \(c_2\), …, \(c_n\). For a fixed \(x\), the determinant of this system is
\[W(x)=\left|\begin{array}{cccc} y_1(x)&y_2(x)&\cdots&y_n(x)\\[4pt] y'_1(x)&y'_2(x)&\cdots&y_n'(x)\\[4pt] \vdots&\vdots&\ddots&\vdots\\[4pt] y_1^{(n-1)}(x)&y_2^{(n-1)}(x)&\cdots&y_n^{(n-1)}(x) \end{array}\right|.\nonumber\]
We call this determinant the Wronskian of \(\{y_1,y_2,\dots,y_n\}\). If \(W(x)\ne0\) for some \(x\) in \((a,b)\) then the system Equation \ref{eq:9.1.13} has only the trivial solution \(c_1=c_2=\cdots=c_n=0\), and Theorem 4.2.2 implies that
\[y=c_1y_1+c_2y_2+\cdots+c_ny_n\nonumber\]
is the general solution of \(Ly=0\) on \((a,b)\).
The next theorem generalizes Theorem 4.1.4.
Suppose the homogeneous linear \(n\)th order equation
\[\label{eq:9.1.14} P_0(x)y^{(n)}+P_1(x)y^{n-1}+\cdots+P_n(x)y=0\]
is normal on \((a,b),\) let \(y_1,\) \(y_2,\) …, \(y_n\) be solutions of Equation \ref{eq:9.1.14} on \((a,b),\) and let \(x_0\) be in \((a,b)\). Then the Wronskian of \(\{y_1,y_2,\dots,y_n\}\) is given by
\[\label{eq:9.1.15} W(x)= W(x_0) \exp\left\{-\int^x_{x_0}{P_1(t) \over P_0(t)} \, dt \right\},\quad a<b.\]
Therefore, either \(W\) has no zeros in \((a,b)\) or \(W\equiv0\) on \((a,b).\)
Formula Equation \ref{eq:9.1.15} is Abel’s formula.
The next theorem is analogous to Theorem 4.1.6.
Suppose \(Ly=0\) is normal on \((a,b)\) and let \(y_1\), \(y_2\), …, \(y_n\) be \(n\) solutions of \(Ly=0\) on \((a,b)\). Then the following statements are equivalent\(;\) that is, they are either all true or all false:
- The general solution of \(Ly=0\) on \((a,b)\) is \(y=c_1y_1+c_2y_2+\cdots+c_ny_n.\)
- \(\{y_1,y_2,\dots,y_n\}\) is a fundamental set of solutions of \(Ly=0\) on \((a,b).\)
- \(\{y_1,y_2,\dots,y_n\}\) is linearly independent on \((a,b).\)
- The Wronskian of \(\{y_1,y_2,\dots,y_n\}\) is nonzero at some point in \((a,b).\)
- The Wronskian of \(\{y_1,y_2,\dots,y_n\}\) is nonzero at all points in \((a,b).\)
In Example 4.2.1 we saw that the solutions \(y_1=x^2\), \(y_2=x^3\), and \(y_3=1/x\) of
\[x^3y'''-x^2y''-2xy'+6y=0 \nonumber\]
are linearly independent on \((-\infty,0)\) and \((0,\infty)\). Calculate the Wronskian of \(\{y_1,y_2,y_3\}\).
Solution
If \(x\ne0\), then
\[W(x)=\left|\begin{array}{ccc}{x^{2}} & {x^{3}} & {\frac{1}{x}} \\ {2 x} & {3 x^{2}} & {-\frac{1}{x^{2}}} \\ {2} & {6 x} & {\frac{2}{x^{3}}}\end{array}\right|=2 x^{3}\left|\begin{array}{ccc}{1} & {x} & {\frac{1}{x^{3}}} \\ {2} & {3 x} & {-\frac{1}{x^{3}}} \\ {1} & {3 x} & {\frac{1}{x^{3}}}\end{array}\right| \nonumber\]
where we factored \(x^2\), \(x\), and \(2\) out of the first, second, and third rows of \(W(x)\), respectively. Adding the second row of the last determinant to the first and third rows yields
\[W(x)=2 x^{3}\left|\begin{array}{ccc}{3} & {4 x} & {0} \\ {2} & {3 x} & {-\frac{1}{x^{3}}} \\ {3} & {6 x} & {0}\end{array}\right|=2 x^{3}\left(\frac{1}{x^{3}}\right)\left|\begin{array}{cc}{3} & {4 x} \\ {3} & {6 x}\end{array}\right|=12 x\nonumber \]
Therefore \(W(x)\ne0\) on \((-\infty,0)\) and \((0,\infty)\).
In Example 4.2.2 we saw that the solutions \(y_1=e^x\), \(y_2=e^{-x}\), \(y_3=e^{2x}\), and \(y_4=e^{-3x}\) of
\[y^{(4)}+y'''-7y''-y'+6y=0 \nonumber\]
are linearly independent on every open interval. Calculate the Wronskian of \(\{y_1,y_2,y_3,y_4\}\).
Solution
For all \(x\),
\[W(x)=\left|\begin{array}{rrrr} e^x&e^{-x}&e^{2x}&e^{-3x}\\ e^x&-e^{-x}&2e^{2x}&-3e^{-3x}\\ e^x&e^{-x}&4e^{2x}&9e^{-3x}\\ e^x&-e^{-x}&8e^{2x}&-27e^{-3x} \end{array}\right|. \nonumber\]
Factoring the exponential common factor from each row yields
\[W(x)=e^{-x}\left|\begin{array}{rrrr}1&1&1&1\\1&-1&2&-3\\1&1&4&9\\ 1&-1&8&-27\end{array}\right|=240e^{-x}, \nonumber\]
from Equation \ref{eq:9.1.12}.
Under the assumptions of Theorem 4.2.4 , it isn’t necessary to obtain a formula for \(W(x)\). Just evaluate \(W(x)\) at a convenient point in \((a, b)\), as we did in Examples 4.2.1 and 4.2.2 .
Suppose \(c\) is in \((a,b)\) and \(\alpha_{1},\) \(\alpha_{2},\) …, are real numbers, not all zero. Under the assumptions of Theorem 10.3.3, suppose \(y_{1}\) and \(y_{2}\) are solutions of Equation 5.1.35 such that
\[\label{eq:9.1.16} \alpha y_{i}(c)+ y_{i}'(c)+\cdots +y_{i}^{(n-1)}(c)=0,\quad 1\le i\le n.\]
Then \(\{y_{1},y_{2},\dots y_{n}\}\) isn’t linearly independent on \((a,b).\)
- Proof
-
Since \(\alpha_{1}\), \(\alpha_{2}\), …, \(\alpha_{n}\) are not all zero, Equation \ref{eq:9.1.14} implies that
\[\left|\begin{array}{ccccccc} y_{1}(c)&y_{1}'(c)&\cdots&y_{1}^{(n-1)}(c)\\ y_{2}(c)&y_{2}'(c)&\cdots&y_{2}^{(n-1)}(c)\\ \vdots&\vdots&\ddots&\vdots\\ y_{n}(c)&y_{n}'(c)&\cdots&y_{n}^{(n-1)}(c)\\ \end{array}\right|=0,\nonumber \]
so
\[\left|\begin{array}{cccccc} y_{1}(c)&y_{2}(c)&\cdots& y_{n}(c)\\ y_{1}'(c)&y_{2}'(c)&\cdots& y_{n}'(c)\\ \vdots&\vdots&\ddots&\vdots\\ y_{1}^{(n-1)}(c)&y_{2}^{(n-1)}(c)(c)&\cdots& y_{n}^{(n-1)}(c)(c)\\ \end{array}\right|=0\nonumber \]
and Theorem 4.2.4 implies the stated conclusion.
General Solution of a Nonhomogeneous Equation
The next theorem shows how to find the general solution of \(Ly=F\) if we know a particular solution of \(Ly=F\) and a fundamental set of solutions of the complementary equation \(Ly=0\).
The probabilities assigned to events by a distribution function on a sample space are given by.
- Proof
-
Suppose \(Ly=F\) is normal on \((a,b).\) Let \(y_p\) be a particular solution of \(Ly=F\) on \((a,b),\) and let \(\{y_1,y_2,\dots,y_n\}\) be a fundamental set of solutions of the complementary equation \(Ly=0\) on \((a,b)\). Then \(y\) is a solution of \(Ly=F\) on \((a,b)\) if and only if
\[y=y_p+c_1y_1+c_2y_2+\cdots+c_ny_n,\nonumber \]
where \(c_1,c_2,\dots,c_n\) are constants.
Suppose for each \(i=1,\) \(2,\) …, \(r\), the function \(y_{p_i}\) is a particular solution of \(Ly=F_i\) on \((a,b).\) Then
\[y_p=y_{p_1}+y_{p_2}+\cdots+y_{p_r} \nonumber\]
is a particular solution of
\[Ly=F_1(x)+F_2(x)+\cdots+F_r(x) \nonumber\]
on \((a,b).\)
We’ll apply Theorems 4.2.6 and 4.2.7 throughout the rest of this chapter.