9.4 Variation of Parameters for Higher Order Equations

Derivation of the method

We assume throughout this section that the nonhomogeneous linear equation

\begin{equation} P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=F(x) \tag{9.4.1}\end{equation}

is normal on an interval \((a,b)\). We’ll abbreviate this equation as \(Ly=F\), where

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

When we speak of solutions of this equation and its complementary 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 \(Ly=F\), 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 \(Ly=F\) in the form

\begin{equation} y_p=u_1y_1+u_2y_2+\cdots+u_ny_n \tag{9.4.2}\end{equation}

where \(\{y_1,y_2,\dots,y_n\}\) is a known fundamental set of solutions of the complementary equation

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

and \(u_1\), \(u_2\), …, \(u_n\) are functions to be determined. We begin by imposing the following \(n-1\) conditions on \(u_1,u_2,\dots,u_n\):

\begin{equation} \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} \tag{9.4.3}\end{equation}

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

\begin{equation} y^{(r)}_p=u_1y^{(r)}_1+u_2y_2^{(r)}\cdots+u_ny^{(r)}_n,\ 0 \le r \le n-1. \tag{9.4.4}\end{equation}

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

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

which reduces to

\[ y_p'=u_1y_1'+u_2y_2'+\cdots+u_ny_n' \]

because of the first equation in (9.4.3). Differentiating this yields

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

which reduces to

\[ y_p''=u_1y_1''+u_2y_2''+\cdots+u_ny_n'' \]

because of the second equation in (9.4.3). Continuing in this way yields (9.4.4).

The last equation in (9.4.4) is

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

Differentiating this yields

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

Substituting this and (9.4.4) into (9.4.1) yields

\[ u_1Ly_1+u_2Ly_2+\cdots+u_nLy_n+P_0(x)\left( u_1'y_1^{(n-1)}+u_2'y_2^{(n-1)}+\cdots+u_n'y_n^{(n-1)}\right)=F(x). \]

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

\[ u_1'y_1^{(n-1)}+u_2'y_2^{(n-1)}+\cdots+u_n'y_n^{(n-1)}={F(x)\over P_0(x)}. \]

Combining this equation with (9.4.3) shows that

\[ y_p=u_1y_1+u_2y_2+\cdots+u_ny_n \]

is a solution of (9.4.1) if

\[ \begin{array}{rcl} u'_1y_1+u'_2y_2+&\cdots&+u'_ny_n=0 \\ u'_1y'_1+u'_2y'_2+&\cdots&+u'_ny'_n=0 \\ \phantom{u'_1y^{(n_1)}+u'_2y_2^{(n-1)}}&\vdots& \phantom{\cdots+u'_ny^{(n-1)}_n=q} \\ u'_1y_1^{(n-2)}+u'_2y^{(n-2)}_2+&\cdots&+u'_ny^{(n-2)}_n =0 \\ u'_1y^{(n-1)}_1+u'_2y^{(n-1)}_2+&\cdots&+u'_n y^{(n-1)}_n=F/P_0, \end{array} \]

which can be written in matrix form as

\begin{equation} \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-1)}&y_2^{(n-1)}&\cdots&y_n^{(n-1)} \end{array} \right] \left[\begin{array}{c}u_1'\\u_2'\\\vdots\\u_{n-1}'\\u_n'\end{array} \right]= \left[\begin{array}{c}0\\0\\ \vdots\\0\\F/ P_0\end{array}\right]. \tag{9.4.5}\end{equation}

The determinant of this system is the Wronskian \(W\) of the fundamental set of solutions \(\{y_1,y_2,\dots,y_n\}\), which has no zeros on \((a,b)\), by Theorem 9.1.4. Solving (9.4.5) by Cramer’s rule yields

\begin{equation} u'_j=(-1)^{n-j}{FW_j\over P_0W},\quad 1\le j\le n, \tag{9.4.6}\end{equation}

where \(W_j\) is the Wronskian of the set of functions obtained by deleting \(y_j\) from \(\{y_1,y_2,\dots,y_n\}\) and keeping the remaining functions in the same order. Equivalently, \(W_j\) is the determinant obtained by deleting the last row and \(j\)-th column of \(W\).

Having obtained \(u_1'\), \(u_2'\)\(, \dots, \)\(u_n'\), we can integrate to obtain \(u_1,\,u_2,\dots,u_n\). As in Section 5.7, 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 \(y_p\).

Remark

For efficiency, it’s best to compute \(W_1\), \(W_2\), …, \(W_n\) first, and then compute \(W\) by expanding in cofactors of the last row; thus,

\[ W=\sum_{j=1}^n(-1)^{n-j}y_j^{(n-1)}W_j. \]

Third Order Equations

If \(n=3\), then

\[ W=\left| \begin{array}{ccc} y_1&y_2&y_3 \\[6pt] y'_1&y'_2&y'_3 \\[6pt] y''_1&y''_2&y''_3 \end{array} \right|. \]

Therefore

\[ W_1=\left| \begin{array}{cc} y_2&y_3 \\[6pt] y'_2&y'_3 \end{array} \right|, \quad W_2=\left| \begin{array}{cc} y_1&y_3 \\[6pt] y'_1&y'_3 \end{array} \right|, \quad W_3=\left| \begin{array}{cc} y_1&y_2 \\[6pt] y'_1&y'_2 \end{array} \right|, \]

and (9.4.6) becomes

\begin{equation} u'_1={FW_1\over P_0W},\quad u'_2=-{FW_2\over P_0W},\quad u'_3={FW_3\over P_0W}. \tag{9.4.7}\end{equation}

Example 9.4.1

Find a particular solution of

\begin{equation} xy'''-y''-xy'+y=8x^2e^x, \tag{9.4.8}\end{equation}

given that \(y_1=x\), \(y_2=e^x\), and \(y_3=e^{-x}\) form a fundamental set of solutions of the complementary equation. Then find the general solution of (9.4.8).

Solution We seek a particular solution of (9.4.8) of the form

\[ y_p=u_1x+u_2e^x+u_3e^{-x}. \]

The Wronskian of \(\{y_1,y_2,y_3\}\) is

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

\begin{eqnarray*} W_1&=& \left| \begin{array}{cr} e^x&e^{-x}\\ e^x&-e^{-x} \end{array} \right|=-2, \\[10pt] W_2&=& \left| \begin{array}{cr} x&e^{-x}\\1&-e^{-x} \end{array} \right|=-e^{-x}(x+1), \\ W_3&=& \left| \begin{array}{cc} x&e^x\\1&e^x \end{array} \right|=e^x(x-1). \end{eqnarray*}

Expanding \(W\) by cofactors of the last row yields

\[ W=0W_1-e^x W_2+e^{-x}W_3=0(-2)-e^x\left(-e^{-x}(x+1)\right) +e^{-x}e^x(x-1)=2x. \]

Since \(F(x)=8x^2e^x\) and \(P_0(x)=x\),

\[ {F\over P_0W}={8x^2e^x\over x\cdot 2x}=4e^x. \]

Therefore, from (9.4.7)

\begin{eqnarray*} u'_1&=&\phantom{-} 4e^xW_1=\phantom{-}4e^x(-2)=-8e^x, \\ u_2'&=&-4e^xW_2=-4e^x\left(-e^{-x}(x+1)\right)=4(x+1), \\ u_3'&=&\phantom{-}4e^xW_3=\phantom{-}4e^x\left(e^x(x-1)\right)=4e^{2x}(x-1). \end{eqnarray*}

Integrating and taking the constants of integration to be zero yields

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

Hence,

\begin{eqnarray*} 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{eqnarray*}

Since \(-e^x\) is a solution of the complementary equation, we redefine

\[ y_p=2xe^x(x-1). \]

Therefore the general solution of (9.4.8) is

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

Fourth Order Equations

If \(n=4\), then

\[ W=\left| \begin{array}{cccc} y_1&y_2&y_3&y_4 \\[6pt] y'_1&y'_2&y'_3&y_4' \\[6pt] y''_1&y''_2&y''_3&y_4''\\[6pt] y'''_1&y'''_2&y'''_3&y_4''' \end{array} \right|, \]

Therefore

\[ W_1=\left| \begin{array}{ccc} y_2&y_3&y_4 \\[6pt] y'_2&y'_3&y_4'\\[6pt] y''_2&y''_3&y_4'' \end{array} \right|, \quad W_2=\left| \begin{array}{ccc} y_1&y_3&y_4 \\[6pt] y'_1&y'_3&y_4'\\[6pt] y''_1&y''_3&y_4'' \end{array} \right|, \]

\[ W_3=\left| \begin{array}{ccc} y_1&y_2&y_4 \\[6pt] y'_1&y'_2&y_4'\\[6pt] y''_1&y''_2&y_4'' \end{array} \right|,\quad W_4=\left| \begin{array}{ccc} y_1&y_2&y_3 \\[6pt] y_1'&y'_2&y_3'\\[6pt] y_1''&y''_2&y_3'' \end{array} \right|, \]

and (9.4.6) becomes

\begin{equation} u'_1=-{FW_1\over P_0W},\quad u'_2={FW_2\over P_0W},\quad u'_3=-{FW_3\over P_0W},\quad u'_4={FW_4\over P_0W}. \tag{9.4.9}\end{equation}

Example 9.4.2

Find a particular solution of

\begin{equation} x^4y^{(4)}+6x^3y'''+2x^2y''-4xy'+4y=12x^2, \tag{9.4.10}\end{equation}

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 complementary equation. Then find the general solution of (9.4.10) on \((-\infty,0)\) and \((0,\infty)\).

Solution We seek a particular solution of (9.4.10) of the form

\[ y_p=u_1x+u_2x^2+{u_3\over x}+{u_4\over x^2}. \]

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

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

\begin{eqnarray*} W_1&=& \left| \begin{array}{ccr} x^2&1/x&1/x^2\\[5pt]2x&-1/x^2&-2/x^3\\[5pt] 2&2/x^3&6/x^4 \end{array} \right|=-{12\over x^4}, \\[10pt] W_2&=& \left| \begin{array}{ccr} x&1/x&1/x^2\\[5pt]1&-1/x^2&-2/x^3\\[5pt] 0&2/x^3&6/x^4 \end{array} \right|=-{6\over x^5}, \\[10pt] W_3&=& \left| \begin{array}{ccc} x&x^2&1/x^2\\[5pt]1&2x&-2/x^3\\[5pt] 0&2&6/x^4 \end{array} \right|={12\over x^2}, \\ W_4&=& \left| \begin{array}{ccc} x&x^2&1/x\\[5pt]1&2x&-1/x^2\\[5pt] 0&2&2/x^3 \end{array} \right|={6\over x}. \end{eqnarray*}

Expanding \(W\) by cofactors of the last row yields

\begin{eqnarray*} W&=&-0W_1+0 W_2-\left(-{6\over x^4}\right)W_3+\left(-{24\over x^5}\right)W_4 \\ &=&{6\over x^4}{12\over x^2}-{24\over x^5}{6\over x}=-{72\over x^6}. \end{eqnarray*}

Since \(F(x)=12x^2\) and \(P_0(x)=x^4\),

\[ {F\over P_0W}={12x^2\over x^4}\left(-{x^6\over72}\right)=-{x^4\over 6}. \]

Therefore, from (9.4.9),

\begin{eqnarray*} u'_1&=&-\left(-{x^4\over6}\right)W_1={x^4\over6}\left(-{12\over x^4}\right)=-2, \\[5pt] u_2'&=&\phantom{-}-{x^4\over6}W_2=-{x^4\over6}\left(-{6\over x^5}\right) ={1\over x}, \\[5pt] u_3'&=&-\left(-{x^4\over6}\right)W_3={x^4\over6}{12\over x^2}=2x^2, \\ u_4'&=&\phantom{-}-{x^4\over6}W_4=-{x^4\over6}{6\over x}=-x^3. \end{eqnarray*}

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

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

Hence,

\begin{eqnarray*} 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{eqnarray*}

Since \(-19x^2/12\) is a solution of the complementary equation, we redefine

\[ y_p=x^2\ln|x|. \]

Therefore

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

is the general solution of (9.4.10) on \((-\infty,0)\) and \((0,\infty)\).

9.4 Exercises

In Exercises 1–21 find a particular solution, given the fundamental set of solutions of the complementary equation.

\(x^3y'''-x^2(x+3)y''+2x(x+3)y'-2(x+3)y=-4x^4\); \(\{x,\,x^2,\,xe^x\}\)

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\(y_p=2x^3\)
\(y'''+6xy''+(6+12x^2)y'+(12x+8x^3)y=x^{1/2}e^{-x^2}\); \(\{e^{-x^2},\,xe^{-x^2},\,x^2e^{-x^2}\}\)

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\(y_p=\dst{8\over 105}x^{7/2} e^{-x^2}\)
\(x^3y'''-3x^2y''+6xy'-6y=2x\); \(\{x,x^2,x^3\}\)

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\(y_p=x\ln|x|\)
\(x^2y'''+2xy''-(x^2+2)y'=2x^2\); \(\{1,\,e^x/x,\,e^{-x}/x\}\)

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\(y_p=\dst{-{2(x^2+2)\over x}}\)
\(x^3y'''-3x^2(x+1)y''+3x(x^2+2x+2)y'-(x^3+3x^2+6x+6)y=x^4e^{-3x}\);

\(\{xe^x,\,x^2e^x,\,x^3e^x\}\)

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\(y_p=\dst{-{xe^{-3x}\over64}}\)
\(x(x^2-2)y'''+(x^2-6)y''+x(2-x^2)y'+(6-x^2)y=2(x^2-2)^2\); \(\{e^x,\,e^{-x},\,1/x\}\)

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\(y_p=-\dst{2x^2\over3}\)
\(xy'''-(x-3)y''-(x+2)y'+(x-1)y=-4e^{-x}\); \(\{e^x,\,e^x/x,\,e^{-x}/x\}\)

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\(y_p=-\dst{e^{-x}(x+1)\over x}\)
\(4x^3y'''+4x^2y''-5xy'+2y=30x^2\); \(\{\sqrt x,\,1/\sqrt x,\,x^2\}\)

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\(y_p=2x^2\ln|x|\)
\(x(x^2-1)y'''+(5x^2+1)y''+2xy'-2y=12x^2\); \(\{x,\,1/(x-1),\,1/(x+1)\}\)

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\(y_p=x^2+1\)
\(x(1-x)y'''+(x^2-3x+3)y''+xy'-y=2(x-1)^2\); \(\{x,\,1/x,e^x/x\}\)

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\(y_p=\dst{2x^2+6\over3}\)
\(x^3y'''+x^2y''-2xy'+2y=x^2\); \(\{x,\,x^2,\,1/x\}\)

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\(y_p=\dst{x^2\ln|x|\over3}\)
\(xy'''-y''-xy'+y=x^2\); \(\{x,\,e^x,\,e^{-x}\}\)

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\(y_p=-x^2-2\)
\(xy^{(4)}+4y'''=6 \ln |x|\); \(\{1,\,x,\,x^2,\,1/x\}\)

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\(\dst{{1\over 4} x^3 \ln |x|-{25\over 48}x^3}\)
\(16x^4y^{(4)}+96x^3y'''+72x^2y''-24xy'+9y=96x^{5/2}\); \(\{\sqrt x,\,1/\sqrt x,\,x^{3/2},\,x^{-3/2}\}\)

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\(y_p=\dst{x^{5/2}\over4}\)
\(x(x^2-6)y^{(4)}+2(x^2-12)y'''+x(6-x^2)y''+2(12-x^2)y'=2(x^2-6)^2\);

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

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\(y_p=\dst{x(12-x^2)\over6}\)
\(x^4y^{(4)}-4x^3y'''+12x^2y''-24xy'+24y=x^4\); \(\{x,\,x^2,\,x^3,\,x^4\}\)

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\(y_p=\dst{x^4\ln|x|\over6}\)
\(x^4y^{(4)}-4x^3y'''+2x^2(6-x^2)y''+4x(x^2-6)y'+(x^4-4x^2+24)y=4x^5e^x\);

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

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\(y_p=\dst{x^3e^x\over2}\)
\(x^4y^{(4)}+6x^3y'''+2x^2y''-4xy'+4y=12x^2\); \(\{x,x^2,1/x,1/x^2\}\)

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\(y_p= x^2\ln|x|\)
\(xy^{(4)}+4y'''-2xy''-4y'+xy=4e^x\); \(\{e^x,\,e^{-x},\,e^x/x,\,e^{-x}/x\}\)

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\(y_p=\dst{xe^x\over2}\)
\(xy^{(4)}+(4-6x)y'''+(13x-18)y''+(26-12x)y'+(4x-12)y=3e^x\); \(\{e^x,\,e^{2x},\,e^x/x,\,e^{2x}/x\}\)

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\(y_p=\dst{3xe^x\over2}\)
\(x^4y^{(4)}-4x^3y'''+x^2(12-x^2)y''+2x(x^2-12)y'+2(12-x^2)y=2x^5\); \(\{x,\,x^2,\,xe^x,\,xe^{-x}\}\)

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\(y_p=-x^3\)

In Exercises 22–33 solve the initial value problem, given the fundamental set of solutions of the complementary equation. Where indicated by C/G , graph the solution.

C/G \(x^3y'''-2x^2y''+3xy'-3y=4x, \quad y(1)=4,\quad y'(1)=4, \quad y''(1)=2\); \(\{x,\,x^3,\,x \ln x\}\)

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\(y=-x(\ln x)^2+3x+x^3-2x\ln x\)
\(x^3y'''-5x^2y''+14xy'-18y=x^3, \quad y(1)=0,\quad y'(1)=1,\quad y''(1)=7\); \(\{x^2,\, x^3,\,x^3 \ln x\}\)

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\(y=\dst{x^3\over 2} (\ln |x|)^2+x^2-x^3+2x^3\ln|x|\)
\(\dst{(5-6x)y'''+(12x-4)y''+(6x-23)y'+(22-12x)y=-(6x-5)^2e^x}\)

\(\dst{y(0)=-4, \quad y'(0)=-{3\over2},\quad y''(0)=-19}\); \(\{e^x,\,e^{2x},\,xe^{-x} \}\)

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\(y=\dst{-{1\over 2}(3x+1)xe^x-3e^x-e^{2x} +4xe^{-x}}\)
\(x^3y'''-6x^2y''+16xy'-16y=9x^4, \quad y(1)=2,\quad y'(1)=1,\quad y''(1)=5\);

\(\{x,\,x^4,\,x^4 \ln |x|\}\)

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\(y=\dst{{3\over 2}x^4(\ln x)^2+3x-x^4+2x^4\ln x}\)
C/G \((x^2-2x+2)y'''-x^2y''+2xy'-2y=(x^2-2x+2)^2, \quad y(0)=0,\quad y'(0)=5\),

\(y''(0)=0\); \(\{x,\,x^2,\,e^x\}\)

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\(y=\dst{-{x^4+12\over6}+3x-x^2+2e^x}\)
\(x^3y'''+x^2y''-2xy'+2y=x(x+1), \quad y(-1)=-6,\quad y'(-1)=\dst{43\over6},\quad y''(-1)= -\dst{5\over2}\);

\(\{x,\,x^2,\,1/x\}\)

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\(y=\dst{\left({x^2\over3}-{x\over2}\right)\ln|x|+4x-2x^2}\)
\((3x-1)y'''-(12x-1)y''+9(x+1)y'-9y=2e^x(3x-1)^2, \quad y(0)=\dst{3\over4}\),

\(y'(0)=\dst{5\over4}, \quad y''(0)=\dst{1\over4}\); \(\{x+1,\,e^x,\,e^{3x}\}\)

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\(y=\dst{-{xe^x(1+3x)\over2}+{x+1\over2}-{e^x\over4}+{e^{3x}\over2}}\)
C/G \((x^2-2)y'''-2xy''+(2-x^2)y'+2xy=2(x^2-2)^2, \quad y(0)=1,\quad y'(0)=-5\),

\(y''(0)=5\); \(\{x^2,\,e^x,\,e^{-x}\}\)

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\(y=-8x+2x^2-2x^3+2e^x-e^{-x}\)
C/G \(x^4y^{(4)}+3x^3y'''-x^2y''+2xy'-2y=9x^2, \quad y(1)=-7,\quad y'(1)= -11,\quad y''(1)=-5\), \(y'''(1)=6; \quad \{x,\,x^2,\,1/x,\,x\ln x\}\)

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\(y=3x^2\ln x-7x^2\)
\((2x-1)y^{(4)}-4xy'''+(5-2x)y''+4xy'-4y=6(2x-1)^2, \quad y(0)=\dst{55\over4}, \quad y'(0)=0\),

\(y''(0)=13, \quad y'''(0)=1\); \(\{x,\,e^x,\,e^{-x},\,e^{2x}\}\)

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\(y=\dst{{3(4x^2+9)\over2}+{x\over2}-{e^x\over2}+{e^{-x}\over2}+{e^{2x}\over4}}\)
\(4x^4y^{(4)}+24x^3y'''+23x^2y''-xy'+y=6x\), \(y(1)=2,\quad y'(1)=0,\quad y''(1)=4,\quad y'''(1)=-\dst{37\over4}\); \(\{x,\sqrt x,1/x,1/\sqrt x\}\)

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\(y=x\ln x+x-\sqrt x+\dst{1\over x}+{1\over\sqrt x}\).
\(x^4y^{{(4)}}+5x^3y'''-3x^2y''-6xy'+6y=40x^3, \quad y(-1)=-1, \; y'(-1)=-7\),

\(y''(-1)=-1,\quad y'''(-1)=-31\); \(\{x,\, x^3,\,1/x,\,1/x^2\}\)

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\(y=\dst{x^3\ln|x|+x-2x^3+{1\over x}-{1\over x^2}}\)
Suppose the equation

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

is normal on an interval \((a,b)\). Let \(\{y_1,y_2,\dots,y_n\}\) be a fundamental set of solutions of its complementary equation on \((a,b)\), let \(W\) be the Wronskian of \(\{y_1,y_2,\dots,y_n\}\), and let \(W_j\) be the determinant obtained by deleting the last row and the \(j\)-th column of \(W\). Suppose \(x_0\) is in \((a,b)\), let

\[ u_j(x)=(-1)^{(n-j)}\int_{x_0}^x{F(t)W_j(t)\over P_0(t)W(t)}\,dt, \quad 1\le j\le n, \]

and define

\[ y_p=u_1y_1+u_2y_2+\cdots+u_ny_n. \]
1. Show that \(y_p\) is a solution of (A) and that
  
  \[ y_p^{(r)}=u_1y^{(r)}_1+u_2y_2^{(r)}\cdots+u_ny^{(r)}_n,\quad 1 \le r \le n-1, \]
  
  and
  
  \[ y_p^{(n)}=u_1y_1^{(n)}+u_2y_2^{(n)}+\cdots+u_ny_n^{(n)}+{F\over P_0}. \]
  
  Hint: See the derivation of the method of variation of parameters at the beginning of the section.
2. Show that \(y_p\) is the solution of the initial value problem
  
  \[ \begin{array}{r} P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=F(x),\\[5pt] \hspace*{20pt} y(x_0)=0,\; y'(x_0)=0,\dots,\quad y^{(n-1)}(x_0)=0. \end{array} \]
3. Show that \(y_p\) can be written as
  
  \[ y_p(x)=\int_{x_0}^x G(x,t)F(t)\,dt, \]
  
  where
  
  \[ G(x,t)={1\over P_0(t)W(t)}\left|\begin{array}{cccc} y_1(t)&y_2(t)&\cdots&y_n(t)\\[5pt] y_1'(t)&y_2'(t)&\cdots&y_n'(t)\\[5pt] \vdots&\vdots&\ddots&\vdots\\[5pt] y_1^{(n-2)}(t)&y_2^{(n-2)}(t)&\cdots&y_n^{(n-2)}(t)\\[5pt] y_1(x)&y_2(x)&\cdots&y_n(x)\end{array}\right|, \]
  
  which is called the Green’s function for (A).
4. Show that
  
  \[ {\partial^{j}G(x,t)\over\partial x^j} ={1\over P_0(t)W(t)}\left|\begin{array}{cccc} y_1(t)&y_2(t)&\cdots&y_n(t)\\[5pt] y_1'(t)&y_2'(t)&\cdots&y_n'(t)\\[5pt] \vdots&\vdots&\ddots&\vdots\\[5pt] y_1^{(n-2)}(t)&y_2^{(n-2)}(t)&\cdots&y_n^{(n-2)}(t)\\[5pt] y_1^{(j)}(x)&y_2^{(j)}(x)&\cdots&y_n^{(j)}(x)\end{array}\right|,\quad 0\le j\le n. \]
5. Show that if \(a<t<b\) then
  
  \[ \left.{\partial^{j}G(x,t)\over\partial x^j}\right|_{x=t}= \left\{\begin{array}{cl} 0,&1\le j\le n-2,\\[5pt] \dst{1\over P_0(t)},&j=n-1. \end{array}\right. \]
6. Show that
  
  \[ y_p^{(j)}(x)=\left\{\begin{array}{cl}\dst{\int_{x_0}^x {\partial^{j}G(x,t)\over\partial x^j}F(t)\,dt},&1\le j\le n-1,\\[10pt] \dst{{F(x)\over P_0(x)}+ \int_{x_0}^x{\partial^{(n)}G(x,t)\over\partial x^n}F(t)\,dt},&j=n. \end{array}\right. \]

In Exercises 35–42 use the method suggested by Exercise 34 to find a particular solution in the form \(y_p=\int_{x_0}^x G(x,t)F(t)\,dt\), given the indicated fundamental set of solutions. Assume that \(x\) and \(x_0\) are in an interval on which the equation is normal.

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

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\(y_p=\dst{\int_{x_0}^x {e^{(x-t)}-3e^{-(x-t)}+2e^{-2(x-t)}\over6} F(t)\,dt}\)
\(x^3y'''+x^2y''-2xy'+2y=F(x); \quad \{x,\,x^2,\,1/x\}\)

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\(y_p=\dst{\int_{x_0}^x {(x-t)^2(2x+t)\over6xt^3} F(t)\,dt}\)
\(x^3y'''-x^2(x+3)y''+2x(x+3)y'-2(x+3)y=F(x); \{x,x^2,xe^x\}\)

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\(y_p=\dst{\int_{x_0}^x {xe^{(x-t)}-x^2+x(t-1)\over t^4} F(t)\,dt}\)
\(x(1-x)y'''+(x^2-3x+3)y''+xy'-y=F(x); \quad \{x,\,1/x,\,e^x/x\}\)

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\(y_p=\dst{\int_{x_0}^x {x^2-t(t-2)-2te^{(x-t)}\over 2x(t-1)^2} F(t)\,dt}\)
\(y^{(4)}-5y''+4y=F(x); \quad \{e^x,\,e^{-x},\,e^{2x},\,e^{-2x}\}\)

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\(y_p=\dst{\int_{x_0}^x {e^{2(x-t)}-2e^{(x-t)}+2e^{-(x-t)}-e^{-2(x-t)}\over12 } F(t)\,dt}\)
\(xy^{(4)}+4y'''=F(x); \quad \{1,\,x,\,x^2,\,1/x\}\)

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\(y_p=\dst{\int_{x_0}^x {(x-t)^3\over6x} F(t)\,dt}\)
\(x^4y^{(4)}+6x^3y'''+2x^2y''-4xy'+4y=F(x)\); \(\{x,x^2,1/x,1/x^2\}\)

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\(y_p=\dst\int_{x_0}^x{(x+t)(x-t)^3\over12x^2t^3}F(t)\,dt\)
\(xy^{(4)}-y'''-4xy'+4y'=F(x); \quad \{1,\,x^2,\,e^{2x}, e^{-2x}\}\)

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\(y_p=\dst{\int_{x_0}^x {e^{2(x-t)}(1+2t)+e^{-2(x-t)}(1-2t)-4x^2+4t^2-2\over32t^2} F(t)\,dt}\)