9.1 Introduction to Linear Higher Order Equations

An \(n\)th order differential equation is said to be linear if it can be written in the form

\begin{equation} y^{(n)}+p_1(x)y^{(n-1)}+\cdots+p_n(x)y=f(x). \tag{9.1.1}\end{equation}

We considered equations of this form with \(n=1\) in Section 2.1 and with \(n=2\) in Chapter 5. In this chapter \(n\) is an arbitrary positive integer.

In this section we sketch the general theory of linear \(n\)th order equations. Since this theory has already been discussed for \(n=2\) in Sections 5.1 and 5.3, we’ll omit proofs.

For convenience, we consider linear differential equations written as

\begin{equation} P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=F(x), \tag{9.1.2}\end{equation}

which can be rewritten as (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 (9.1.2) by \(Ly\); that is,

\[ Ly=P_0y^{(n)}+P_1y^{(n-1)}+\cdots+P_ny. \]

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 (9.1.1) with \(p_1\), …, \(p_n\) and \(f\) continuous on \((a,b)\).

The next theorem is analogous to Theorem 5.3.1.

Theorem 9.1.1

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

\[ Ly=F, \quad y(x_0)=k_0,\quad y'(x_0)=k_1,\dots,\quad y^{(n-1)}(x_0)=k_{n-1} \]

has a unique solution on \((a,b).\)

Homogeneous Equations

Eqn. (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

\begin{equation} y=c_1y_1+c_2y_2+\cdots+c_ny_n \tag{9.1.3}\end{equation}

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 (9.1.3). In this case we say that (9.1.3) is the general solution of \(Ly=0\) on \((a,b)\).

It can be shown (Exercises 14 and 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

\begin{equation} c_1y_1(x)+c_2y_2(x)+\cdots+c_ny_n(x)=0,\quad a<x<b, \tag{9.1.4}\end{equation}

are \(c_1=c_2=\cdots=c_n=0\). If (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 5.1.3.

Theorem 9.1.2

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’s linearly independent on \((a,b)\).

Example 9.1.1

The equation

\begin{equation} x^3y'''-x^2y''-2xy'+6y=0 \tag{9.1.5}\end{equation}

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 (9.1.5) on \((-\infty, 0)\) and \((0,\infty)\).

Solution Suppose

\begin{equation} c_1x^2+c_2x^3+{c_3\over x}=0 \tag{9.1.6}\end{equation}

on \((0,\infty)\). We must show that \(c_1=c_2=c_3=0\). Differentiating (9.1.6) twice yields the system

\begin{equation} \begin{array}{rcl} c_1x^2+c_2x^3+\dst{c_3\over x}&=&0\\[6pt] 2c_1x+3c_2x^2-\dst{c_3\over x^2}&=&0\\[6pt] 2c_1+6c_2x+\dst{2c_3\over x^3}&=&0. \end{array} \tag{9.1.7}\end{equation}

If (9.1.7) holds for all \(x\) in \((0,\infty)\), then it certainly holds at \(x=1\); therefore,

\begin{equation} \begin{array}{rcl} \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} \tag{9.1.8}\end{equation}

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’s 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 (9.1.8) is

\[ \left|\begin{array}{rrr}1&1&1\\2&3&-1\\2&6&2\end{array}\right|= \left|\begin{array}{rrr}1&0&0\\2&1&-3\\2&4&0\end{array}\right|=12, \]

it follows that (9.1.8) has only the trivial solution, so \(\{y_1,y_2,y_3\}\) is linearly independent on \((0,\infty)\). Now Theorem 9.1.2 implies that

\[ y=c_1x^2+c_2x^3+{c_3\over x} \]

is the general solution of (9.1.5) on \((0,\infty)\). To see that this is also true on \((-\infty,0)\), assume that (9.1.6) holds on \((-\infty,0)\). Setting \(x=-1\) in (9.1.7) yields

\begin{eqnarray*} \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{eqnarray*}

Since the determinant of this system is

\[ \left|\begin{array}{rrr}1&-1&-1\\-2&3&-1\\2&-6&-2\end{array}\right|= \left|\begin{array}{rrr}1&0&0\\-2&1&-3\\2&-4&0\end{array}\right|=-12, \]

it follows that \(c_1=c_2=c_3=0\); that is, \(\{y_1,y_2,y_3\}\) is linearly independent on \((-\infty,0)\).

Example 9.1.2

The equation

\begin{equation} y^{(4)}+y'''-7y''-y'+6y=0 \tag{9.1.9}\end{equation}

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 (9.1.9).

Solution Suppose \(c_1\), \(c_2\), \(c_3\), and \(c_4\) are constants such that

\begin{equation} c_1e^x+c_2e^{-x}+c_3e^{2x}+c_4e^{-3x}=0 \tag{9.1.10}\end{equation}

for all \(x\). We must show that \(c_1=c_2=c_3=c_4=0\). Differentiating (9.1.10) three times yields the system

\begin{equation} \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} \tag{9.1.11}\end{equation}

If (9.1.11) holds for all \(x\), then it certainly holds for \(x=0\). Therefore

\[ \begin{array}{rcl} c_1+c_2+\phantom{2}c_3+\phantom{27}c_4&=&0\\ c_1-c_2+2c_3-\phantom{2}3c_4&=&0\\ c_1+c_2+4c_3+\phantom{2}9c_4&=&0\\ c_1-c_2+8c_3-27c_4&=&0. \end{array} \]

The determinant of this system is

\begin{equation} \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|\\[24pt]&=& \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} \tag{9.1.12}\end{equation}

so the system has only the trivial solution \(c_1=c_2=c_3=c_4=0\). Now Theorem 9.1.2 implies that

\[ y=c_1e^x+c_2e^{-x}+c_3e^{2x}+c_4e^{-3x} \]

is the general solution of (9.1.9).

The Wronskian

We can use the method used in Examples 9.1.1 and 9.1.2 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

\[ c_1y_1+c_2y_2+\cdots+c_ny_n=0,\quad a<x<b, \]

then differentiating \(n-1\) times leads to the \(n\times n\) system of equations

\begin{equation} \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} \tag{9.1.13}\end{equation}

for \(c_1\), \(c_2\), …, \(c_n\). For a fixed \(x\), the determinant of this system is

\[ W(x)=\left|\begin{array}{cccc} y_1(x)&y_2(x)&\cdots&y_n(x)\\[6pt] y'_1(x)&y'_2(x)&\cdots&y_n'(x)\\[6pt] \vdots&\vdots&\ddots&\vdots\\[6pt] y_1^{(n-1)}(x)&y_2^{(n-1)}(x)&\cdots&y_n^{(n-1)}(x) \end{array}\right|. \]

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 (9.1.13) has only the trivial solution \(c_1=c_2=\cdots=c_n=0\), and Theorem 9.1.2 implies that

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

is the general solution of \(Ly=0\) on \((a,b)\).

The next theorem generalizes Theorem 5.1.4. The proof is sketched in (Exercises 17–20).

Theorem 9.1.3

Suppose the homogeneous linear \(n\)th order equation

\begin{equation} P_0(x)y^{(n)}+P_1(x)y^{n-1}+\cdots+P_n(x)y=0 \tag{9.1.14}\end{equation}

is normal on \((a,b),\) let \(y_1,\) \(y_2,\) …, \(y_n\) be solutions of (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

\begin{equation} W(x)=W(x_0)\exp\left\{-\int^x_{x_0}{P_1(t)\over P_0(t)}\, dt\right\},\quad a<x<b. \tag{9.1.15}\end{equation}

Therefore\(,\) either \(W\) has no zeros in \((a,b)\) or \(W\equiv0\) on \((a,b).\)

Formula (9.1.15) is Abel’s formula.

The next theorem is analogous to Theorem 5.1.6..

Theorem 9.1.4

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 \((a,b).\)

Example 9.1.3

In Example 9.1.1 we saw that the solutions \(y_1=x^2\), \(y_2=x^3\), and \(y_3=1/x\) of

\[ x^3y'''-x^2y''-2xy'+6y=0 \]

are linearly independent on \((-\infty,0)\) and \((0,\infty)\). Calculate the Wronskian of \(\{y_1,y_2,y_3\}\).

Solution If \(x\ne0\), then

\[ W(x)=\left|\begin{array}{ccc} x^2&\phantom{2}x^3&\phantom{-}\dst{1\over x}\\[6pt] 2x&3x^2\phantom{^2}&\dst-\dst{1\over x^2}\\[6pt] 2&6x\phantom{x^2}&\phantom{-}\dst{2\over x^3}\end{array} \right| =2x^3\left|\begin{array}{ccc} 1&\phantom{3}x&\phantom{-}\dst{1\over x^3}\\[6pt]2&3x&-\dst{1\over x^3} \\[6pt]1&3x&\phantom{-}\dst{1\over x^3}\end{array}\right|, \]

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

\[ W(x)=2x^3\left|\begin{array}{ccc} 3&4x&\phantom{-}0\\[6pt]2&3x&-\dst{1\over x^3} \\[6pt]3&6x&\phantom{-}0\end{array}\right| =2x^3\left(1\over x^3\right)\left|\begin{array}{cc}3&4x\\3&6x \end{array}\right|=12x. \]

Therefore \(W(x)\ne0\) on \((-\infty,0)\) and \((0,\infty)\).

Example 9.1.4

In Example 9.1.2 we saw that the solutions \(y_1=e^x\), \(y_2=e^{-x}\), \(y_3=e^{2x}\), and \(y_4=e^{-3x}\) of

\[ y^{(4)}+y'''-7y''-y'+6y=0 \]

are linearly independent on every open interval. Calculate the Wronskian of \(\{y_1,y_2,y_3,y_4\}\).

Solution For all \(x\),

\[ W(x)=\left|\begin{array}{rrrr} e^x&e^{-x}&e^{2x}&e^{-3x}\\ e^x&-e^{-x}&2e^{2x}&-3e^{-3x}\\ e^x&e^{-x}&4e^{2x}&9e^{-3x}\\ e^x&-e^{-x}&8e^{2x}&-27e^{-3x} \end{array}\right|. \]

Factoring the exponential common factor from each row yields

\[ W(x)=e^{-x}\left|\begin{array}{rrrr}1&1&1&1\\1&-1&2&-3\\1&1&4&9\\ 1&-1&8&-27\end{array}\right|=240e^{-x}, \]

from (9.1.12).

Remark

Under the assumptions of Theorem 9.1.4, 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 9.1.1 and 9.1.2.

Theorem 9.1.5

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 (5.1.35) such that

\begin{equation} \alpha y_{i}(c)+ y_{i}'(c)+\cdots +y_{i}^{(n-1)}(c)=0,\quad 1\le i\le n. \tag{9.1.16}\end{equation}

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, (9.1.14) implies that

\[ \left|\begin{array}{ccccccc} y_{1}(c)&y_{1}'(c)&\cdots&y_{1}^{(n-1)}(c)\\ y_{2}(c)&y_{2}'(c)&\cdots&y_{2}^{(n-1)}(c)\\ \vdots&\vdots&\ddots&\vdots\\ y_{n}(c)&y_{n}'(c)&\cdots&y_{n}^{(n-1)}(c)\\ \end{array}\right|=0, \]

\[ \left|\begin{array}{cccccc} y_{1}(c)&y_{2}(c)&\cdots& y_{n}(c)\\ y_{1}'(c)&y_{2}'(c)&\cdots& y_{n}'(c)\\ \vdots&\vdots&\ddots&\vdots\\ y_{1}^{(n-1)}(c)&y_{2}^{(n-1)}(c)(c)&\cdots& y_{n}^{(n-1)}(c)(c)\\ \end{array}\right|=0 \]

and Theorem 9.1.4 implies the stated conclusion.

General Solution of a Nonhomogeneous Equation

The next theorem is analogous to Theorem 5.3.2. It 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 \(Ly=0\).

Theorem 9.1.6

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

\[ y=y_p+c_1y_1+c_2y_2+\cdots+c_ny_n, \]

where \(c_1,c_2,\dots,c_n\) are constants.

The next theorem is analogous to Theorem 5.3.2.

Theorem 9.1.7

Suppose for each \(i=1,\) \(2,\) …, \(r\), the function \(y_{p_i}\) is a particular solution of \(Ly=F_i\) on \((a,b).\) Then

\[ y_p=y_{p_1}+y_{p_2}+\cdots+y_{p_r} \]

is a particular solution of

\[ Ly=F_1(x)+F_2(x)+\cdots+F_r(x) \]

on \((a,b).\)

We’ll apply Theorems 9.1.6 and 9.1.7 throughout the rest of this chapter.

9.1 Exercises

Verify that the given function is the solution of the initial value problem.

(a) \( x^3y'''-3x^2y''+6xy'-6y=-\dst{24\over x}, \quad y(-1)=0\), \( y'(-1)=0, \quad y''(-1)=0\);

\(\dst{y=-6x-8x^2-3x^3+{1\over x}}\)

(b) \(\dst{ y'''-{1\over x}y''-y'+{1\over x}y={x^2-4\over x^4}, \quad y(1)={3\over2},\quad y'(1)={1\over2}}\), \( y''(1)=1\);

\(y=x+\dst{1\over2x}\)

(c) \(xy'''-y''-xy'+y=x^2, \quad y(1)=2,\quad y'(1)=5,\quad y''(1)=-1\);

\(y=-x^2-2+2e^{(x-1)}-e^{-(x-1)}+4x\)

(d) \(4x^3y'''+4x^2y''-5xy'+2y=30x^2, \quad y(1)=5,\quad y'(1)=\dst{17\over2}\);

\(y''(1)=\dst{63\over4};\quad y=2x^2\ln x-x^{1/2}+2x^{-1/2}+4x^2\)

(e) \(x^4y^{(4)}-4x^3y'''+12x^2y''-24xy'+24y=6x^4, \quad y(1)=-2\),

\(y'(1)=-9, \quad y''(1)=-27,\quad y'''(1)=-52\);

\(y=x^4\ln x+x-2x^2+3x^3-4x^4\)

(f) \(xy^{(4)}-y'''-4xy''+4y'=96x^2, \quad y(1)=-5,\quad y'(1)=-24\)

\(y''(1)=-36; \quad y'''(1)=-48;\quad y=9-12x+6x^2-8x^3\)
Solve the initial value problem

\[ x^3y'''-x^2 y''-2xy'+6y=0, \quad y(-1)=-4, \quad y'(-1)=-14,\quad y''(-1)=-20. \]

Hint: See Example \(\ref{example:9.1.1}\).

Show answer

\(y=\dst{2x^2-3x^3+{1\over x}}\)
Solve the initial value problem

\[ y^{(4)}+y'''-7y''-y'+6y=0, \quad y(0)=5,\quad y'(0)=-6,\quad y''(0)=10,\quad y'''(0)-36. \]

Hint: See Example \(\ref{example:9.1.2}\).

Show answer

\(y=2e^x+3e^{-x}-e^{2x}+e^{-3x}\)
Find solutions \(y_1\), \(y_2\), …, \(y_n\) of the equation \(y^{(n)}=0\) that satisfy the initial conditions

\[ y_i^{(j)}(x_0)=\left\{\begin{array}{cl} 0,&j\ne i-1,\\[3pt] 1,&j=i-1,\end{array}\right.\; 1\le i\le n. \]

Show answer

\(\dst{y_i={(x-x_0)^{i-1}\over(i-1)!},\,1\le i \le n}\)
1. Verify that the function
  
  \[ y=c_1x^3+c_2x^2+{c_3\over x} \]
  
  satisfies
  
  \[ x^3 y'''-x^2y''-2xy'+6y=0 \tag*{\rm (A)} \]
  
  if \(c_1\), \(c_2\), and \(c_3\) are constants.
2. Use (a) to find solutions \(y_1\), \(y_2\), and \(y_3\) of (A) such that
  
  \[ \begin{array}{rl} y_1(1)&=1,\quad y_1'(1)=0,\quad y_1''(1)=0 \\[3pt] y_2(1)&=0,\quad y_2'(1)=1,\quad y_2''(1)=0 \\[3pt] y_3(1)&=0,\quad y_3'(1)=0,\quad y_3''(1)=1. \end{array} \]
3. Use (b) to find the solution of (A) such that
  
  \[ y(1)=k_0,\quad y'(1)=k_1,\quad y''(1)=k_2. \]
Show answer

(b) \(\dst{ y_1=-{1\over2}x^3+x^2+{1\over2x}, \quad y_2={1\over3}x^2-{1\over3x},\quad y_3={1\over4} x^3-{1\over 3}x^2+{1\over12x}}\)

(c) \(y=k_0y_1+k_1y_2+k_2 y_3\)
Verify that the given functions are solutions of the given equation, and show that they form a fundamental set of solutions of the equation on any interval on which the equation is normal.

(a) \(y'''+y''-y'-y=0; \quad\{e^x,\,e^{-x},\,xe^{-x}\}\)

(b) \(y'''-3y''+7y'-5y=0; \quad\{e^x,\,e^x\cos2x,\,e^x\sin2x\}\).

(c) \(xy'''-y''-xy'+y=0; \quad \{e^x,\,e^{-x},\,x\}\)

(d) \(x^2y'''+2xy''-(x^2+2)y=0; \quad\dst\{e^x/ x,\,e^{-x}/ x,\,1\}\)

(e) \((x^2-2x+2)y'''-x^2y''+2xy'-2y=0; \quad \{x,\,x^2,\,e^x\}\)

(f) \((2x-1)y^{(4)}-4xy'''+(5-2x)y''+4xy'-4y=0; \quad\{x,\,e^x,\,e^{-x},e^{2x}\}\)

(g) \(xy^{(4)}-y'''-4xy'+4y'=0; \quad\{1,x^2,\,e^{2x},\,e^{-2x}\}\)
Find the Wronskian \(W\) of a set of three solutions of

\[ y'''+2xy''+e^xy'-y=0, \]

given that \(W(0)=2\).

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\(2e^{-x^2}\)
Find the Wronskian \(W\) of a set of four solutions of

\[ y^{(4)}+(\tan x)y'''+x^2y''+2xy=0, \]

given that \(W(\pi/4)=K\).

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\(\sqrt{2} K\cos x\)
1. Evaluate the Wronskian \(W\) \(\{e^x,\,xe^x,\, x^2e^x\}\). Evaluate \(W(0)\).
2. Verify that \(y_1\), \(y_2\), and \(y_3\) satisfy
  
  \[ y'''-3y''+3y'-y=0. \tag*{\rm (A)} \]
3. Use \(W(0)\) from (a) and Abel’s formula to calculate \(W(x)\).
4. What is the general solution of (A)?
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(a) \(W(x)=2e^{3x}\) (d) \(y=e^x(c_1+c_2x+c_3x^2)\)
Compute the Wronskian of the given set of functions.

(a) \(\{1,\,e^x,\,e^{-x}\}\) (b) \(\{e^x,\, e^x\sin x,\,e^x\cos x\}\)

(c) \(\{2,\,x+1,\,x^2+2\}\) (d) \( x,\,x\ln x,\,1/x\}\)

(e) \(\dst{\{1,\,x,\,{x^2\over2!},\, {x^3\over3!}\,,\cdots,\,{x^n\over n!}}\}\) (f) \(\{e^x,\,e^{-x},\,x\}\)

(g) \(\{e^x/x,\,e^{-x}/x,\,1\}\) (h) \(\{x,\,x^2,\,e^x\}\)

(i) \(\{x,\,x^3,\,1/x,\,1/x^2\}\) (j) \(\{e^x,\,e^{-x},\,x,\,e^{2x}\}\)

(k) \(\{e^{2x},\,e^{-2x},\,1,\,x^2\}\)

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(a) \(2\) (b) \(-e^{3x}\)(c) \(4\) (d) \(4/x^2\) (e) \(1\) (f) \(2x\) (g) \(2/x^2\)(h) \(e^x(x^2-2x+2)\)

(i) \(-240/x^5\) (j) \(6e^{2x}(2x-1)\)(l) \(-128x\)
Suppose \(Ly=0\) is normal on \((a,b)\) and \(x_0\) is in \((a,b)\). Use Theorem 9.1.1 to show that \(y\equiv0\) is the only solution of the initial value problem

\[ Ly=0, \quad y(x_0)=0,\quad y'(x_0)=0,\dots, y^{(n-1)}(x_0)=0, \]

on \((a,b)\).
Prove: If \(y_1\), \(y_2\), …, \(y_n\) are solutions of \(Ly=0\) and the functions

\[ z_i=\sum^n_{j=1}a_{ij}y_j,\quad 1\le i\le n, \]

form a fundamental set of solutions of \(Ly=0\), then so do \(y_1\), \(y_2\), …, \(y_n\).
Prove: If

\[ y=c_1y_1+c_2y_2+\cdots+c_ky_k+y_p \]

is a solution of a linear equation \(Ly=F\) for every choice of the constants \(c_1\), \(c_2\) ,…, \(c_k\), then \(Ly_i=0\) for \(1\le i\le k\).
Suppose \(Ly=0\) is normal on \((a,b)\) and let \(x_0\) be in \((a,b)\). For \(1\le i\le n\), let \(y_i\) be the solution of the initial value problem

\[ Ly_i=0, \quad y_i^{(j)} (x_0)= \left\{\begin{array}{cl} 0,& j\ne i-1,\\ [3pt] 1,&j=i-1,\end{array}\right. 1\le i\le n, \]

where \(x_0\) is an arbitrary point in \((a,b)\). Show that any solution of \(Ly=0\) on \((a, b)\), can be written as

\[ y=c_1y_1+c_2y_2+\cdots+c_ny_n, \]

with \(c_j=y^{(j-1)}(x_0)\).
Suppose \(\{y_1, y_2,\dots, y_n\}\) is a fundamental set of solutions of

\[ \dst P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=0 \]

on \((a,b)\), and let

\[ \begin{array}{rl} z_1&=a_{11}y_1+a_{12}y_2+\cdots+a_{1n}y_n\\ z_2&=a_{21}y_1+a_{22}y_2+\cdots+a_{2n}y_n\\ \phantom{z_1}&\vdots\phantom{_1y_1+a}\vdots \phantom{_2y_2+\cdots+a}\vdots\phantom{_ny_n} \phantom{=b}\vdots\\ z_n&=a_{n1}y_1+a_{n2}y_2+\cdots+a_{nn}y_n, \end{array} \]

where the \(\{a_{ij}\}\) are constants. Show that \(\{z_1, z_2,\dots, z_n\}\) is a fundamental set of solutions of (A) if and only if the determinant

\[ \left|\begin{array}{cccc} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}\end{array}\right| \]

is nonzero.Hint: The determinant of a product of \(n\times n\) matrices equals the product of the determinants.
Show that \(\{y_1,y_2,\dots,y_n\}\) is linearly dependent on \((a,b)\) if and only if at least one of the functions \(y_1\), \(y_2\), …, \(y_n\) can be written as a linear combination of the others on \((a,b)\).

Take the following as a hint in Exercises 17–19:

By the definition of determinant,

\[ \left|\begin{array}{cccc} a_{11}&a_{12}&\cdots&a_{1n}\\[6pt] a_{21}&a_{22}&\cdots&a_{2n}\\[6pt] \vdots&\vdots&\ddots&\vdots\\[6pt] a_{n1}&a_{n2}&\cdots&a_{nn}\end{array}\right| =\sum\pm a_{1i_1}a_{2i_2},\dots,a_{ni_n}, \]

where the sum is over all permutations \((i_1,i_2,\dots,i_n)\) of \((1,2,\dots, n)\) and the choice of \(+\) or \(-\) in each term depends only on the permutation associated with that term.

Prove: If

\[ A(u_1,u_2,\dots,u_n)= \left|\begin{array}{cccc} a_{11}&a_{12}&\cdots&a_{1n}\\[6pt] a_{21}&a_{22}&\cdots&a_{2n}\\[6pt] \vdots&\vdots&\ddots&\vdots\\[6pt] a_{n-1,1}&a_{n-1,2}&\cdots&a_{n-1,n}\\[6pt] u_1&u_2&\cdots&u_n\end{array}\right|, \]

then

\[ A(u_1+v_1, u_2+v_2,\dots, u_n+v_n)=A(u_1,u_2,\dots,u_n)+A(v_1,v_2,\dots, v_n). \]
Let

\[ F=\left|\begin{array}{cccc} f_{11}&f_{12}&\cdots&f_{1n}\\[6pt] f_{21}&f_{22}&\cdots&f_{2n}\\[6pt] \vdots&\vdots&\ddots&\vdots\\[6pt] f_{n1}&f_{n2}&\cdots&f_{nn}\end{array}\right|, \]

where \(f_{ij}\; (1\le i,\; j\le n)\) is differentiable. Show that

\[ F'=F_1+F_2+\cdots+F_n, \]

where \(F_i\) is the determinant obtained by differentiating the \(i\)th row of \(F\).
Use Exercise 18 to show that if \(W\) is the Wronskian of the \(n\)-times differentiable functions \(y_1\), \(y_2\), …, \(y_n\), then

\[ W'= \left|\begin{array}{cccc} y_1&y_2&\cdots&y_n\\[6pt] y'_1&y'_2&\cdots&y'_n\\[6pt] \vdots&\vdots&\ddots&\vdots\\[6pt] y_1^{(n-2)}&y_2^{(n-2)}&\cdots&y_n^{(n-2)}\\[6pt] y_1^{(n)}&y_2^{(n)}&\cdots&y_n^{(n)} \end{array}\right|. \]
Use Exercises 17 and 19 to show that if \(W\) is the Wronskian of solutions \(\{y_1,y_2,\dots,y_n\}\) of the normal equation

\[ P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=0, \tag*{\rm (A)} \]

then \(W'=-P_1W/P_0\). Derive Abel’s formula (Eqn. (9.1.15)) from this. Hint: Use (A) to write \(y^{(n)}\) in terms of \(y,y',\dots,y^{(n-1)}\).
Prove Theorem 9.1.6.
Prove Theorem 9.1.7.
Show that if the Wronskian of the \(n\)-times continuously differentiable functions \(\{y_1,y_2,\dots,y_n\}\) has no zeros in \((a,b)\), then the differential equation obtained by expanding the determinant

\[ \left|\begin{array}{ccccc} y&y_1&y_2&\cdots&y_n\\[6pt] y'&y'_1&y'_2&\cdots&y'_n\\[6pt] \vdots&\vdots&\vdots&\ddots& \vdots\\[6pt] y^{(n)}&y_{1}^{(n)}&y_2^{(n)}&\cdots&y_n^{(n)} \end{array}\right|=0, \]

in cofactors of its first column is normal and has \(\{y_1,y_2,\dots,y_n\}\) as a fundamental set of solutions on \((a,b)\).
Use the method suggested by Exercise 23 to find a linear homogeneous equation such that the given set of functions is a fundamental set of solutions on intervals on which the Wronskian of the set has no zeros.

(a) \(\{x,\,x^2-1,\,x^2+1\}\) (b) \(\{e^x,\,e^{-x},\,x\}\)

(c) \(\{e^x,\,xe^{-x},\,1\}\) (d) \(\{x,\,x^2,\,e^x\}\)

(e) \(\{x,\,x^2,\,1/x\}\) (f) \(\{x+1,\,e^x,\,e^{3x}\}\)

(g) \(\{x,\,x^3,\,1/x,\,1/x^2\}\) (h) \(\{x,\,x\ln x,\,1/x,\,x^2\}\)

(i) \(\{e^x,\,e^{-x},\,x,\,e^{2x}\}\) (j) \(\{e^{2x},\,e^{-2x},\,1,\,x^2\}\)

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(a) \(y'''=0\) (b) \(xy'''-y''-xy'+y=0\) (c) \((2x-3)y'''-2y''-(2x-5)y'=0\)

(d) \((x^2-2x+2)y'''-x^2y''+2xy'-2y=0\) (e) \(x^3y'''+x^2y''-2xy'+2y=0\)

(f) \((3x-1)y'''-(12x-1)y''+9(x+1)y'-9y=0\)

(g) \(x^4y^{(4)}+5x^3y'''-3x^2y''-6xy'+6y=0\)

(h) \(x^4y^{(4)}+3x^2y'''-x^2y''+2xy'-2y=0\)

(i) \((2x-1)y^{(4)}-4xy'''+(5-2x)y''+4xy'-4y=0\)

(j) \(xy^{(4)}-y'''-4xy''+4y'=0\)