9.3 Undetermined Coefficients for Higher Order Equations

In this section we consider the constant coefficient equation

\begin{equation} a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=F(x), \tag{9.3.1}\end{equation}

where \(n\ge3\) and \(F\) is a linear combination of functions of the form

\[ e^{\alpha x}\left(p_0+p_1x+\cdots+p_kx^k\right) \]

\[ e^{\lambda x}\left[\left(p_0+p_1x+\cdots+p_kx^k\right) \cos\omega x+ \left(q_0+q_1x+\cdots+q_kx^k\right) \sin\omega x\right]. \]

From Theorem 9.1.5, the general solution of (9.3.1) is \(y=y_p+y_c\), where \(y_p\) is a particular solution of (9.3.1) and \(y_c\) is the general solution of the complementary equation

\[ a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=0. \]

In Section 9.2 we learned how to find \(y_c\). Here we will learn how to find \(y_p\) when the forcing function has the form stated above. The procedure that we use is a generalization of the method that we used in Sections 5.4 and 5.5, and is again called method of undetermined coefficients. Since the underlying ideas are the same as those in Sections 5.4 and 5.5, we’ll give an informal presentation based on examples.

Forcing Functions of the Form \(e^{\alpha x}\left(p_0+p_1x+\cdots+p_kx^k\right)\)

We first consider equations of the form

\[ a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=e^{\alpha x}\left(p_0+p_1x+\cdots+p_kx^k\right). \]

Example 9.3.1

Find a particular solution of

\begin{equation} y'''+3y''+2y'-y=e^x(21+24x+28x^2+5x^3). \tag{9.3.2}\end{equation}

Solution Substituting

\begin{eqnarray*} y&=&ue^x, \\ y'&=&e^x(u'+u), \\ y''&=&e^x(u''+2u'+u), \\ y'''&=&e^x(u'''+3u''+3u'+u) \end{eqnarray*}

into (9.3.2) and canceling \(e^x\) yields

\[ (u'''+3u''+3u'+u)+3(u''+2u'+u)+2(u'+u)-u =21+24x+28x^2+5x^3, \]

\begin{equation} u'''+6u''+11u'+5u=21+24x+28x^2+5x^3. \tag{9.3.3}\end{equation}

Since the unknown \(u\) appears on the left, we can see that (9.3.3) has a particular solution of the form

\[ u_p=A+Bx+Cx^2+Dx^3. \]

Then

\begin{eqnarray*} u_p'&=&B+2Cx+3Dx^2 \\ u_p''&=&2C+6Dx \\ u_p'''&=&6D. \end{eqnarray*}

Substituting from the last four equations into the left side of (9.3.3) yields

\begin{eqnarray*} u_p'''+6u_p''+11u_p'+5u_p&=&6D+6(2C+6Dx)+11(B+2Cx+3Dx^2) \\ &&+5(A+Bx+Cx^2+Dx^3) \\ &=&(5A+11B+12C+6D)+(5B+22C+36D)x \\ &&+(5C+33D)x^2+5Dx^3. \end{eqnarray*}

Comparing coefficients of like powers of \(x\) on the right sides of this equation and (9.3.3) shows that \(u_p\) satisfies (9.3.3) if

\[ \begin{array}{rcr} 5D&=&5\phantom{.}\\ 5C+33D&=&28\phantom{.}\\ 5B+22C+36D&=&24\phantom{.}\\ 5A+11B+12C+\phantom{3}6D&=&21. \end{array} \]

Solving these equations successively yields \(D=1\), \(C=-1\), \(B=2\), \(A=1\). Therefore

\[ u_p=1+2x-x^2+x^3 \]

is a particular solution of (9.3.3), so

\[ y_p=e^xu_p=e^x(1+2x-x^2+x^3) \]

is a particular solution of (9.3.2) (Figure 9.3.1).

Example 9.3.2

Find a particular solution of

\begin{equation} y^{(4)}-y'''-6y''+4y'+8y=e^{2x}(4+19x+6x^2). \tag{9.3.4}\end{equation}

Solution Substituting

\begin{eqnarray*} y&=&ue^{2x}, \\ y'&=&e^{2x}(u'+2u), \\ y''&=&e^{2x}(u''+4u'+4u), \\ y'''&=&e^{2x}(u'''+6u''+12u'+8u), \\ y^{(4)}&=&e^{2x}(u^{(4)}+8u'''+24u''+32u'+16u) \end{eqnarray*}

into (9.3.4) and canceling \(e^{2x}\) yields

\begin{eqnarray*} &&(u^{(4)}+8u'''+24u''+32u'+16u)-(u'''+6u''+12u'+8u) \\ &&-6(u''+4u'+4u)+4(u'+2u)+8u=4+19x+6x^2, \end{eqnarray*}

\begin{equation} u^{(4)}+7u'''+12u''=4+19x+6x^2. \tag{9.3.5}\end{equation}

Since neither \(u\) nor \(u'\) appear on the left, we can see that (9.3.5) has a particular solution of the form

\begin{equation} u_p=Ax^2+Bx^3+Cx^4. \tag{9.3.6}\end{equation}

Then

\begin{eqnarray*} u_p'&=&2Ax+3Bx^2+4Cx^3 \\ u_p''&=&2A+6Bx+12Cx^2 \\ u_p'''&=&6B+24Cx \\ u_p^{(4)}&=&24C. \end{eqnarray*}

Substituting \(u_p''\), \(u_p'''\), and \(u_p^{(4)}\) into the left side of (9.3.5) yields

\begin{eqnarray*} u_p^{(4)}+7u_p'''+12u_p''&=&24C+7(6B+24Cx)+12(2A+6Bx+12Cx^2) \\ &=&(24A+42B+24C)+(72B+168C)x+144Cx^2. \end{eqnarray*}

Comparing coefficients of like powers of \(x\) on the right sides of this equation and (9.3.5) shows that \(u_p\) satisfies (9.3.5) if

\[ \begin{array}{rcr} 144C&=&6\phantom{.}\\ 72B+168C&=&19\phantom{.}\\ 24A+42B+\phantom{1}24C&=&4. \end{array} \]

Solving these equations successively yields \(C=1/24\), \(B=1/6\), \(A=-1/6\). Substituting these into (9.3.6) shows that

\[ u_p={x^2\over24}(-4+4x+x^2) \]

is a particular solution of (9.3.5), so

\[ y_p=e^{2x}u_p={x^2e^{2x}\over24}(-4+4x+x^2) \]

is a particular solution of (9.3.4). (Figure 9.3.2).

y_p=x^2e^2x24(-4+4x+x^2) — Figure 9.3.2. \(y_p=\dst{{x^2e^{2x}\over24}(-4+4x+x^2)}\)

Forcing Functions of the Form \(e^{\lambda x}\left(P(x)\cos\omega x+Q(x)\sin\omega x\right)\)

We now consider equations of the form

\[ a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny= e^{\lambda x}\left(P(x)\cos\omega x+Q(x)\sin\omega x\right), \]

where \(P\) and \(Q\) are polynomials.

Example 9.3.3

Find a particular solution of

\begin{equation} y'''+y''-4y'-4y=e^x[(5-5x)\cos x+(2+5x)\sin x]. \tag{9.3.7}\end{equation}

Solution Substituting

\begin{eqnarray*} y&=&ue^x, \\ y'&=&e^x(u'+u), \\ y''&=&e^x(u''+2u'+u), \\ y'''&=&e^x(u'''+3u''+3u'+u) \end{eqnarray*}

into (9.3.7) and canceling \(e^x\) yields

\[ (u'''+3u''+3u'+u)+(u''+2u'+u)-4(u'+u)-4u =(5-5x)\cos x+(2+5x)\sin x, \]

\begin{equation} u'''+4u''+u'-6u=(5-5x)\cos x+(2+5x)\sin x. \tag{9.3.8}\end{equation}

Since \(\cos x\) and \(\sin x\) are not solutions of the complementary equation

\[ u'''+4u''+u'-6u=0, \]

a theorem analogous to Theorem 5.5.1 implies that (9.3.8) has a particular solution of the form

\begin{equation} u_p=(A_0+A_1x)\cos x+(B_0+B_1x)\sin x. \tag{9.3.9}\end{equation}

Then

\begin{eqnarray*} u_p'=(A_1+B_0+B_1x)\cos x+(B_1-A_0-A_1x)\sin x, \\ u_p''=(2B_1-A_0-A_1x)\cos x-(2A_1+B_0+B_1x)\sin x, \\ u_p'''=-(3A_1+B_0+B_1x)\cos x-(3B_1-A_0-A_1x)\sin x,\end{eqnarray*}

\[ \begin{array}{rcl} u_p'''+4u_p''+u_p'-6u_p&=&-\left[10A_0+2A_1-8B_1+10A_1x\right]\cos x\\ &&- \left[10B_0+2B_1+8A_1+10B_1x\right]\sin x. \end{array} \]

Comparing the coefficients of \(x\cos x\), \(x\sin x\), \(\cos x\), and \(\sin x\) here with the corresponding coefficients in (9.3.8) shows that \(u_p\) is a solution of (9.3.8) if

\[ \begin{array}{rcr} -10A_1&=&-5\phantom{.}\\ -10B_1&=&\phantom{-}5\phantom{.}\\ -10A_0-2A_1+8B_1&=&5\phantom{.}\\ -10B_0-2B_1-8A_1&=&2. \end{array} \]

Solving the first two equations yields \(A_1=1/2\), \(B_1=-1/2\). Substituting these into the last two equations yields

\begin{eqnarray*} -10A_0&=&5+2A_1-8B_1=10\phantom{.} \\ -10B_0&=&2+2B_1+8A_1=5, \end{eqnarray*}

so \(A_0=-1\), \(B_0=-1/2\). Substituting \(A_0=-1\), \(A_1=1/2\), \(B_0=-1/2\), \(B_1=-1/2\) into (9.3.9) shows that

\[ u_p=-\dst{1\over2}\left[(2-x)\cos x+(1+x)\sin x\right] \]

is a particular solution of (9.3.8), so

\[ y_p=e^xu_p=-{e^x\over2}\left[(2-x)\cos x+(1+x)\sin x\right] \]

is a particular solution of (9.3.7) (Figure 9.3.3).

y_p=e^xu_p=-e^x2[(2-x)x+(1+x)x] — Figure 9.3.3. \(y_p=\dst{e^xu_p=-{e^x\over2}\left[(2-x)\cos x+(1+x)\sin x\right]}\)

Example 9.3.4

Find a particular solution of

\begin{equation} y'''+4y''+6y'+4y= e^{-x}\left[(1-6x)\cos x-(3+2x)\sin x\right]. \tag{9.3.10}\end{equation}

Solution Substituting

\begin{eqnarray*} y&=&ue^{-x}, \\ y'&=&e^{-x}(u'-u), \\ y''&=&e^{-x}(u''-2u'+u), \\ y'''&=&e^{-x}(u'''-3u''+3u'-u) \end{eqnarray*}

into (9.3.10) and canceling \(e^{-x}\) yields

\[ (u'''-3u''+3u'-u)+4(u''-2u'+u)+6(u'-u)+4u =(1-6x)\cos x-(3+2x)\sin x, \]

\begin{equation} u'''+u''+u'+u=(1-6x)\cos x-(3+2x)\sin x. \tag{9.3.11}\end{equation}

Since \(\cos x\) and \(\sin x\) are solutions of the complementary equation

\[ u'''+u''+u'+u=0, \]

a theorem analogous to Theorem 5.5.1 implies that (9.3.11) has a particular solution of the form

\begin{equation} u_p=(A_0x+A_1x^2)\cos x+(B_0x+B_1x^2)\sin x. \tag{9.3.12}\end{equation}

Then

\begin{eqnarray*} u_p'&=&[A_0+(2A_1+B_0)x+B_1x^2]\cos x+[B_0+(2B_1-A_0)x-A_1x^2]\sin x, \\ u_p''&=&[2A_1+2B_0-(A_0-4B_1)x-A_1x^2]\cos x \\ &&+ [2B_1-2A_0-(B_0+4A_1)x-B_1x^2]\sin x, \\ u_p'''&=&-[3A_0-6B_1+(6A_1+B_0)x+B_1x^2]\cos x \\ &&-[3B_0+6A_1+(6B_1-A_0)x-A_1x^2]\sin x, \end{eqnarray*}

\[ \begin{array}{rcl} u_p'''+u_p''+u_p'+u_p&=& -[2A_0-2B_0-2A_1-6B_1+(4A_1-4B_1)x]\cos x\\ &&-[2B_0+2A_0-2B_1+6A_1+(4B_1+4A_1)x]\sin x. \end{array} \]

Comparing the coefficients of \(x\cos x\), \(x\sin x\), \(\cos x\), and \(\sin x\) here with the corresponding coefficients in (9.3.11) shows that \(u_p\) is a solution of (9.3.11) if

\[ \begin{array}{rcr} -4A_1+4B_1&=&-6\phantom{.}\\ -4A_1-4B_1&=&-2\phantom{.}\\ -2A_0+2B_0+2A_1+6B_1&=&\phantom{-}1\phantom{.}\\ -2A_0-2B_0-6A_1+2B_1&=&-3. \end{array} \]

Solving the first two equations yields \(A_1=1\), \(B_1=-1/2\). Substituting these into the last two equations yields

\begin{eqnarray*} -2A_0+2B_0&=&\phantom{-}1-2A_1-6B_1=2\phantom{.} \\ -2A_0-2B_0&=&-3+6A_1-2B_1=4, \end{eqnarray*}

y_p=-xe^-x2[(3-2x)x+(1+x)x] — Figure 9.3.4. \(\dst{y_p=-{xe^{-x}\over2}\left[(3-2x)\cos x+(1+x)\sin x\right]}\)

so \(A_0=-3/2\) and \(B_0=-1/2\). Substituting \(A_0=-3/2\), \(A_1=1\), \(B_0=-1/2\), \(B_1=-1/2\) into (9.3.12) shows that

\[ u_p=-\dst{x\over2}\left[(3-2x)\cos x+(1+x)\sin x\right] \]

is a particular solution of (9.3.11), so

\[ y_p=e^{-x}u_p=-{xe^{-x}\over2}\left[(3-2x)\cos x+(1+x)\sin x\right] \]

(Figure 9.3.4) is a particular solution of (9.3.10).

9.3 Exercises

In Exercises 1–59 find a particular solution.

\(y'''-6y''+11y'-6y=-e^{-x}(4+76x-24x^2)\)

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\(y_p=e^{-x}(2+x-x^2)\)
\(y'''-2y''-5y'+6y=e^{-3x}(32-23x+6x^2)\)

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\(y_p=-\dst{e^{-3x}\over4}(3-x+x^2)\)
\(4y'''+8y''-y'-2y=-e^x(4+45x+9x^2)\)

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\(y_p=e^x(1+x-x^2)\)
\(y'''+3y''-y'-3y=e^{-2x}(2-17x+3x^2)\)

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\(y_p=e^{-2x}(1-5x+x^2)\).
\(y'''+3y''-y'-3y=e^x(-1+2x+24x^2+16x^3)\)

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\(y_p=-\dst{xe^x\over2}(1-x+x^2-x^3)\)
\(y'''+y''-2y=e^x(14+34x+15x^2)\)

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\(y_p=x^2e^x(1+x)\)
\(4y'''+8y''-y'-2y=-e^{-2x}(1-15x)\)

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\(y_p=\dst{xe^{-2x}\over2}(2+x)\)
\(y'''-y''-y'+y=e^x(7+6x)\)

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\(y_p=\dst{x^2e^x\over2}(2+x)\)
\(2y'''-7y''+4y'+4y=e^{2x}(17+30x)\)

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\(y_p=\dst{x^2e^{2x}\over2}(1+2x)\)
\(y'''-5y''+3y'+9y=2e^{3x}(11-24x^2)\)

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\(y_p=x^2e^{3x}(2+x-x^2)\)
\(y'''-7y''+8y'+16y=2e^{4x}(13+15x)\)

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\(y_p=x^2e^{4x}(2+x)\)
\(8y'''-12y''+6y'-y=e^{x/2}(1+4x)\)

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\(y_p=\dst{x^3e^{x/2}\over48}(1+x)\)
\(y^{(4)}+3y'''-3y''-7y'+6y=-e^{-x}(12+8x-8x^2)\)

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\(y_p=e^{-x}(1-2x+x^2)\)
\(y^{(4)}+3y'''+y''-3y'-2y=-3e^{2x}(11+12x)\)

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\(y_p=e^{2x}(1-x)\)
\(y^{(4)}+8y'''+24y''+32y'=-16e^{-2x}(1+x+x^2-x^3)\)

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\(y_p=e^{-2x}(1+x+x^2-x^3)\)
\(4y^{(4)}-11y''-9y'-2y=-e^x(1-6x)\)

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

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\(y_p=e^x(1+x)^2\)
\(y^{(4)}-4y'''+6y''-4y'+2y=e^{2x}(24+x+x^4)\)

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\(y_p=xe^x(1+x^3)\)
\(2y^{(4)}+5y'''-5y'-2y=18e^x(5+2x)\)

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\(y_p=xe^x(2+x)\)
\(y^{(4)}+y'''-2y''-6y'-4y=-e^{2x}(4+28x+15x^2)\)

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\(y_p=\dst{xe^{2x}\over6}(1-x^2)\)
\(2y^{(4)}+y'''-2y'-y=3e^{-x/2}(1-6x)\)

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\(y_p=4xe^{-x/2}(1+x)\)
\(y^{(4)}-5y''+4y=e^x(3+x-3x^2)\)

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\(y_p=\dst{xe^x\over6}(1+x^2)\)
\(y^{(4)}-2y'''-3y''+4y'+4y=e^{2x}(13+33x+18x^2)\)

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\(y_p=\dst{x^2e^{2x}\over6}(1+x+x^2)\)
\(y^{(4)}-3y'''+4y'=e^{2x}(15+26x+12x^2)\)

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\(y_p=\dst{x^2e^{2x}\over6}(3+x+x^2)\)
\(y^{(4)}-2y'''+2y'-y=e^x(1+x)\)

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\(y_p=\dst{x^3e^x\over48}(2+x)\)
\(2y^{(4)}-5y'''+3y''+y'-y=e^x(11+12x)\)

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\(y_p=\dst{x^3e^x\over6}(1+x)\)
\(y^{(4)}+3y'''+3y''+y'=e^{-x}(5-24x+10x^2)\)

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\(y_p=-\dst{x^3e^{-x}\over6}(1-x+x^2)\)
\(y^{(4)}-7y'''+18y''-20y'+8y=e^{2x}(3-8x-5x^2)\)

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\(y_p=\dst{x^3e^{2x}\over12}(2+x-x^2)\)
\(y'''-y''-4y'+4y=e^{-x}\left[(16+10x)\cos x+(30-10x)\sin x\right]\)

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\(y_p=e^{-x}\left[(1+x)\cos x+(2-x)\sin x\right]\)
\(y'''+y''-4y'-4y=e^{-x}\left[(1-22x)\cos 2x-(1+6x)\sin2x\right]\)

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\(y_p=e^{-x}\left[(1-x)\cos2x+(1+x)\sin2 x\right]\)
\(y'''-y''+2y'-2y=e^{2x}[(27+5x-x^2)\cos x+(2+13x+9x^2)\sin x]\)

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\(y_p=e^{2x}[(1+x-x^2)\cos x+(1+2x)\sin x]\)
\(y'''-2y''+y'-2y=-e^x[(9-5x+4x^2)\cos 2x-(6-5x-3x^2)\sin2x]\)

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\(y_p=\dst{e^x\over2}[(1+x)\cos 2x+(1-x+x^2)\sin2x]\)
\(y'''+3y''+4y'+12y=8\cos2x-16\sin2x\)

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\(y_p=\dst{x\over13}(8\cos2x+14\sin2x)\)
\(y'''-y''+2y=e^x[(20+4x)\cos x-(12+12x)\sin x]\)

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\(y_p=xe^x[(1+x)\cos x+(3+x)\sin x]\)
\(y'''-7y''+20y'-24y=-e^{2x}[(13-8x)\cos 2x-(8-4x)\sin2x]\)

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\(y_p=\dst{xe^{2x}\over2}[(3-x)\cos 2x+\sin2x]\)
\(y'''-6y''+18y'=-e^{3x}[(2-3x)\cos 3x-(3+3x)\sin3x]\)

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\(y_p=-\dst{xe^{3x}\over12}(x\cos 3x+\sin3x)\)
\(y^{(4)}+2y'''-2y''-8y'-8y=e^x(8\cos x+16\sin x)\)

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\(y_p=-\dst{e^x\over10}(\cos x+7\sin x)\)
\(y^{(4)}-3y'''+2y''+2y'-4y=e^x(2\cos2x -\sin2x)\)

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\(y_p=\dst{e^x\over12}(\cos2x-\sin2x)\)
\(y^{(4)}-8y'''+24y''-32y'+15y=e^{2x}(15x\cos2x+32\sin2x)\)

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\(y_p=xe^{2x}\cos2x\)
\(y^{(4)}+6y'''+13y''+12y'+4y=e^{-x}[(4-x)\cos x-(5+x)\sin x]\)

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\(y_p=-\dst{e^{-x}\over2}[(1+x)\cos x+(2-x)\sin x]\)
\(y^{(4)}+3y'''+2y''-2y'-4y=-e^{-x} (\cos x-\sin x)\)

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\(y_p=\dst{xe^{-x}\over10}(\cos x+2\sin x)\)
\(y^{(4)}-5y'''+13y''-19y'+10y=e^x (\cos2x+\sin2x)\)

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\(y_p=\dst{xe^x\over40}(3\cos2x-\sin2x)\)
\(y^{(4)}+8y'''+32y''+64y'+39y=e^{-2x}[(4-15x)\cos3x-(4+15x)\sin 3x]\)

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\(y_p=\dst{xe^{-2x}\over8}[(1-x)\cos 3x+(1+x)\sin3x]\)
\(y^{(4)}-5y'''+13y''-19y'+10y=e^x[(7+8x)\cos 2x+(8-4x)\sin2x]\)

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\(y_p=-\dst{xe^x\over4}(1+x)\sin2x\)
\(y^{(4)}+4y'''+8y''+8y'+4y=-2e^{-x} (\cos x-2\sin x)\)

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\(y_p=\dst{x^2e^{-x}\over4}(\cos x-2\sin x)\)
\(y^{(4)}-8y'''+32y''-64y'+64y=e^{2x} (\cos2x-\sin2x)\)

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\(y_p=-\dst{x^2e^{2x}\over32}(\cos2x-\sin2x)\)
\(y^{(4)}-8y'''+26y''-40y'+25y=e^{2x}[3\cos x-(1+3x)\sin x]\)

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\(y_p=\dst{x^2e^{2x}\over8}(1+x)\sin x\)
\(y'''-4y''+5y'-2y=e^{2x}-4e^x-2\cos x+4\sin x\)

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\(y_p=2x^2e^x+xe^{2x}-\cos x\)
\(y'''-y''+y'-y=5e^{2x}+2e^x-4\cos x+4\sin x\)

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\(y_p=e^{2x}+xe^x+2x\cos x\)
\(y'''-y'=-2(1+x)+4e^x-6e^{-x}+96e^{3x}\)

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\(y_p=2x+x^2+2xe^x-3xe^{-x}+4e^{3x}\)
\(y'''-4y''+9y'-10y=10e^{2x}+20e^x\sin2x-10\)

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\(y_p=xe^x(\cos2x-2\sin2x)+2xe^{2x}+1\)
\(y'''+3y''+3y'+y=12e^{-x}+9\cos2x-13\sin2x\)

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\(y_p=x^2e^{-2x}(1+2x)-\cos2x+\sin2x\)
\(y'''+y''-y'-y=4e^{-x}(1-6x)-2x\cos x+2(1+x)\sin x \)

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\(y_p=2x^2(1+x)e^{-x}+x\cos x-2\sin x\)
\(y^{(4)}-5y''+4y=-12e^x+6e^{-x}+10\cos x\)

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\(y_p=2xe^x+xe^{-x}+\cos x\)
\(y^{(4)}-4y'''+11y''-14y'+10y=-e^x(\sin x+2\cos2x)\)

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\(y_p=\dst{xe^x\over6}(\cos x+\sin2x)\)
\(y^{(4)}+2y'''-3y''-4y'+4y=2e^x(1+x)+e^{-2x}\)

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\(y_p=\dst{x^2\over54}[(2+2x)e^x+3e^{-2x}]\)
\(y^{(4)}+4y=\sinh x\cos x-\cosh x\sin x\)

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\(y_p=\dst{x\over8}\sinh x\sin x\)
\(y^{(4)}+5y'''+9y''+7y'+2y=e^{-x}(30+24x)-e^{-2x}\)

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\(y_p=x^3(1+x)e^{-x}+xe^{-2x}\)
\(y^{(4)}-4y'''+7y''-6y'+2y=e^x(12x-2\cos x+2\sin x)\)

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\(y_p=xe^x(2x^2+\cos x+\sin x)\)

In Exercises 60–68 find the general solution.

\(y'''-y''-y'+y=e^{2x}(10+3x)\)

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\(y=e^{2x}(1+x)+c_1e^{-x}+e^x(c_2+c_3x)\)
\(y'''+y''-2y=-e^{3x}(9+67x+17x^2)\)

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\(y=e^{3x}\dst\left(1-x-{x^2\over2}\right)+ c_1e^x+e^{-x}(c_2\cos x+c_3\sin x)\)
\(y'''-6y''+11y'-6y=e^{2x}(5-4x-3x^2)\)

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\(y=xe^{2x}(1+x)^2+c_1e^x+c_2e^{2x}+c_3e^{3x}\)
\(y'''+2y''+y'=-2e^{-x}(7-18x+6x^2)\)

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\(y=x^2e^{-x}(1-x)^2+c_1 +e^{-x}(c_2+c_3x)\)
\(y'''-3y''+3y'-y=e^x(1+x)\)

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\(y=\dst{x^3e^x\over24}(4+x) +e^x(c_1+c_2x+c_3x^2)\)
\(y^{(4)}-2y''+y=-e^{-x}(4-9x+3x^2)\)

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\(y=\dst{x^2e^{-x}\over16}(1+2x-x^2) +e^x(c_1+c_2x)+e^{-x}(c_3+c_4x)\)
\(y'''+2y''-y'-2y=e^{-2x}\left[(23-2x)\cos x+(8-9x)\sin x\right]\)

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\(y=e^{-2x}\dst{\left[\left(1+{x\over2}\right) \cos x+\left({3\over2}-2x\right)\sin x\right]} +c_1e^x+c_2e^{-x}+c_3e^{-2x}\)
\(y^{(4)}-3y'''+4y''-2y'=e^x\left[(28+6x)\cos 2x+(11-12x)\sin2x\right]\)

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\(y=-xe^x\sin2x+c_1+c_2e^x +e^x(c_3\cos x+c_4\sin x)\)
\(y^{(4)}-4y'''+14y''-20y'+25y=e^x\left[(2+6x)\cos 2x+3\sin2x\right]\)

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\(y=-\dst{x^2e^x\over16}(1+x)\cos2x +e^x\left[(c_1+c_2x)\cos2x+(c_3+c_4x)\sin2x\right]\)

In Exercises 69–74 solve the initial value problem and graph the solution.

C/G \(y'''-2y''-5y'+6y=2e^x(1-6x),\quad y(0)=2, \quad y'(0)=7,\quad y''(0)=9\)

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\(y=(x^2+2)e^x-e^{-2x}+e^{3x}\)
C/G \(y'''-y''-y'+y=-e^{-x}(4-8x),\quad y(0)=2, \quad y'(0)=0,\quad y''(0)=0\)

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\(y=e^{-x}(1+x+x^2) +(1-x)e^x\)
C/G \(4y'''-3y'-y=e^{-x/2}(2-3x),\quad y(0)=-1, \quad y'(0)=15,\quad y''(0)=-17\)

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\(y=\dst\left({x^2\over12}+16\right)xe^{-x/2} -e^x\)
C/G \(y^{(4)}+2y'''+2y''+2y'+y=e^{-x}(20-12x),\, y(0)=3,\; y'(0)=-4,\; y''(0)=7,\; y'''(0)=-22\)

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\(y=(2-x)(x^2+1)e^{-x}+\cos x-\sin x\)
C/G \(y'''+2y''+y'+2y=30\cos x-10\sin x, \quad y(0)=3,\quad y'(0)=-4,\quad y''(0)=16\)

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\(y=(2-x)\cos x-(1-7x)\sin x+e^{-2x}\)
C/G \(y^{(4)}-3y'''+5y''-2y'=-2e^x(\cos x-\sin x),\; y(0)=2,\; y'(0)=0,\; y''(0)~=~-1, \\ \hspace*{30pt} y'''(0)=-5\)

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\(2 +e^x\left[(1+x)\cos x-\sin x-1\right]\)
Prove: A function \(y\) is a solution of the constant coefficient nonhomogeneous equation

\[ a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=e^{\alpha x}G(x) \tag*{\rm(A)} \]

if and only if \(y=ue^{\alpha x}\), where \(u\) satisfies the differential equation

\[ a_0u^{(n)}+{p^{(n-1)}(\alpha)\over(n-1)!}u^{(n-1)}+ {p^{(n-2)}(\alpha)\over(n-2)!}u^{(n-2)}+\cdots+p(\alpha)u=G(x) \tag*{\rm(B)} \]

and

\[ p(r)=a_0r^n+a_1r^{n-1} + \cdots + a_n \]

is the characteristic polynomial of the complementary equation

\[ a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=0. \]
Prove:
1. The equation
  
  \[ \begin{array}{lcl} a_0u^{(n)}&+&\dst{p^{(n-1)}(\alpha)\over(n-1)!}u^{(n-1)}+ \dst{p^{(n-2)}(\alpha)\over(n-2)!}u^{(n-2)}+\cdots+p(\alpha)u\\ &=&\left(p_0+p_1x+\cdots+p_kx^k\right)\cos \omega x\\&&\,+ \left(q_0+q_1x+\cdots+q_kx^k\right)\sin\omega x \end{array} \tag*{\rm(A)} \]
  
  has a particular solution of the form
  
  \[ u_p=x^m\left(u_0+u_1x+\cdots+u_kx^k\right)\cos\omega x+ \left(v_0+v_1x+\cdots+v_kx^k\right)\sin\omega x. \]
2. If \(\lambda+i\omega\) is a zero of \(p\) with multiplicity \(m\ge1\), then (A) can be written as
  
  \[ a(u''+\omega^2 u)= \left(p_0+p_1x+\cdots+p_kx^k\right)\cos\omega x+ \left(q_0+q_1x+\cdots+q_kx^k\right)\sin\omega x, \]
  
  which has a particular solution of the form
  
  \[ u_p=U(x)\cos\omega x+V(x)\sin\omega x, \]
  
  where
  
  \[ U(x)=u_0x+u_1x^2+\cdots+u_kx^{k+1},\,V(x)=v_0x+v_1x^2+\cdots+v_kx^{k+1} \]
  
  and
  
  \[ \begin{array}{rcl} a(U''(x)+2\omega V'(x))&=&p_0+p_1x+\cdots+p_kx^k\\[10pt] a(V''(x)-2\omega U'(x))&=&q_0+q_1x+\cdots+q_kx^k. \end{array} \]