2.3E: Exercises
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- 17367
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In Exercises \((2.3E.1)\) to \((2.3E.6)\), find a particular solution by the method used in Example \((2.3.2)\). Then find the general solution and, where indicated, solve the initial value problem and graph the solution.
Exercise \(\PageIndex{1}\)
\(y''+5y'-6y=22+18x-18x^2\)
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Exercise \(\PageIndex{2}\)
\(y''-4y'+5y=1+5x\)
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Exercise \(\PageIndex{3}\)
\(y''+8y'+7y=-8-x+24x^2+7x^3\)
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Exercise \(\PageIndex{4}\)
\(y''-4y'+4y=2+8x-4x^2\)
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Exercise \(\PageIndex{5}\)
\(y''+2y'+10y=4+26x+6x^2+10x^3, \quad y(0)=2, \quad y'(0)=9\)
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Exercise \(\PageIndex{6}\)
\(y''+6y'+10y=22+20x, \quad y(0)=2,\; y'(0)=-2\)
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Exercise \(\PageIndex{7}\)
Show that the method used in Example \((2.3.2)\) won't yield a particular solution of
\begin{equation}\label{eq:2.3E.1}
y''+y'=1+2x+x^2;
\end{equation}
that is, \eqref{eq:2.3E.1} doesn't have a particular solution of the form \(y_p=A+Bx+Cx^2\), where \(A\), \(B\), and \(C\) are constants.
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In Exercises \((2.3E.8)\) to \((2.3E.13)\), find a particular solution by the method used in Example \((2.3.3)\).
Exercise \(\PageIndex{8}\)
\(x^2y''+7xy'+8y=\displaystyle{6\over x}\)
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Exercise \(\PageIndex{9}\)
\(x^2y''-7xy'+7y=13x^{1/2}\)
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Exercise \(\PageIndex{10}\)
\(x^2y''-xy'+y=2x^3\)
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Exercise \(\PageIndex{11}\)
\(x^2y''+5xy'+4y=\displaystyle{1\over x^3}\)
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Exercise \(\PageIndex{12}\)
\(x^2y''+xy'+y=10x^{1/3}\)
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Exercise \(\PageIndex{13}\)
\(x^2y''-3xy'+13y=2x^4\)
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Exercise \(\PageIndex{14}\)
Show that the method suggested for finding a particular solution in Exercises \((2.3E.8)\) to \((2.3E.13)\) won't yield a particular solution of
\begin{equation}\label{eq:2.3E.2}
x^2y''+3xy'-3y={1\over x^3};
\end{equation}
that is, \eqref{eq:2.3E.2} doesn't have a particular solution of the form \(y_p=A/x^3\).
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Exercise \(\PageIndex{15}\)
Prove: If \(a\), \(b\), \(c\), \(\alpha\), and \(M\) are constants and \(M\ne0\) then
\begin{eqnarray*}
ax^2y''+bxy'+cy=M x^\alpha
\end{eqnarray*}
has a particular solution \(y_p=Ax^\alpha\) (\(A=\) constant) if and only if \(a\alpha(\alpha-1)+b\alpha+c\ne0\).
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If \(a\), \(b\), \(c\), and \(\alpha\) are constants, then
\begin{eqnarray*}
a(e^{\alpha x})''+b(e^{\alpha x})'+ce^{\alpha x}=(a\alpha^2+b\alpha+c)e^{\alpha x}.
\end{eqnarray*}
Use this in Exercises \((2.3E.16)\) to \((2.3E.21)\) to find a particular solution. Then find the general solution and, where indicated, solve the initial value problem and graph the solution.
Exercise \(\PageIndex{16}\)
\(y''+5y'-6y=6e^{3x}\)
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Exercise \(\PageIndex{17}\)
\(y''-4y'+5y=e^{2x}\)
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Exercise \(\PageIndex{18}\)
\(y''+8y'+7y=10e^{-2x}, \quad y(0)=-2,\; y'(0)=10\)
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Exercise \(\PageIndex{19}\)
\(y''-4y'+4y=e^{x}, \quad y(0)=2,\quad y'(0)=0\)
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Exercise \(\PageIndex{20}\)
\(y''+2y'+10y=e^{x/2}\)
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Exercise \(\PageIndex{21}\)
\(y''+6y'+10y=e^{-3x}\)
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Exercise \(\PageIndex{22}\)
Show that the method suggested for finding a particular solution in Exercises \((2.3E.16)\) to \((2.3E.21)\) won't yield a particular solution of
\begin{equation}\label{eq:2.3E.3}
y''-7y'+12y=5e^{4x};
\end{equation}
that is, \eqref{eq:2.3E.3} doesn't have a particular solution of the form \(y_p=Ae^{4x}\).
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Exercise \(\PageIndex{23}\)
Prove: If \(\alpha\) and \(M\) are constants and \(M\ne0\) then constant coefficient equation
\begin{eqnarray*}
ay''+by'+cy=M e^{\alpha x}
\end{eqnarray*}
has a particular solution \(y_p=Ae^{\alpha x}\) (\(A=\) constant) if and only if \(e^{\alpha x}\) isn't a solution of the complementary equation.
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If \(\omega\) is a constant, differentiating a linear combination of \(\cos\omega x\) and \(\sin\omega x\) with respect to \(x\) yields another linear combination of \(\cos\omega x\) and \(\sin\omega x\). In Exercises \((2.3E.24)\) to \((2.3E.29)\) use this to find a particular solution of the equation. Then find the general solution and, where indicated, solve the initial value problem and graph the solution.
Exercise \(\PageIndex{24}\)
\(y''-8y'+16y=23\cos x-7\sin x\)
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Exercise \(\PageIndex{25}\)
\(y''+y'=-8\cos2x+6\sin2x\)
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Exercise \(\PageIndex{26}\)
\(y''-2y'+3y=-6\cos3x+6\sin3x\)
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Exercise \(\PageIndex{27}\)
\(y''+6y'+13y=18\cos x+6\sin x \)
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Exercise \(\PageIndex{28}\)
\(y''+7y'+12y=-2\cos2x+36\sin2x, \quad y(0)=-3,\quad y'(0)=3\)
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Exercise \(\PageIndex{29}\)
\(y''-6y'+9y=18\cos3x+18\sin3x, \quad y(0)=2,\quad y'(0)=2\)
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Exercise \(\PageIndex{30}\)
Find the general solution of
\begin{eqnarray*}
y''+\omega_0^2y =M\cos\omega x+N\sin\omega x,
\end{eqnarray*}
where \(M\) and \(N\) are constants and \(\omega\) and \(\omega_0\) are distinct positive numbers.
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Exercise \(\PageIndex{31}\)
Show that the method suggested for finding a particular solution in Exercises \((2.3E.24)\) to \((2.3E.29)\) won't yield a particular solution of
\begin{equation}\label{eq:2.3E.4}
y''+y=\cos x+\sin x;
\end{equation}
that is, \eqref{eq:2.3E.4} does not have a particular solution of the form \(y_p=A\cos x+B\sin x\).
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Exercise \(\PageIndex{32}\)
Prove: If \(M\), \(N\) are constants (not both zero) and \(\omega>0\), the constant coefficient equation
\begin{equation}\label{eq:2.3E.5}
ay''+by'+cy=M\cos\omega x+N\sin\omega x
\end{equation}
has a particular solution that's a linear combination of \(\cos\omega x\) and \(\sin\omega x\) if and only if the left side of \eqref{eq:2.3E.5} is not of the form \(a(y''+\omega^2y)\), so that \(\cos\omega x\) and \(\sin\omega x\) are solutions of the complementary equation.
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In Exercises \((2.3E.33)\) to \(2.3E.38)\), refer to the cited exercises and use the principal of superposition to find a particular solution. Then find the general solution.
Exercise \(\PageIndex{33}\)
\(y''+5y'-6y=22+18x-18x^2+6e^{3x}\)
(See Exercises \((2.3E.1)\) and \((2.3E.16)\).)
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Exercise \(\PageIndex{34}\)
\(y''-4y'+5y=1+5x+e^{2x}\)
(See Exercises \(2.3E.2)\) and \((2.3E.17)\).)
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Exercise \(\PageIndex{35}\)
\(y''+8y'+7y=-8-x+24x^2+7x^3+10e^{-2x}\)
(See Exercises \((2.3E.3)\) and \((2.3E.18)\).)
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Exercise \(\PageIndex{36}\)
\(y''-4y'+4y=2+8x-4x^2+e^{x}\)
(See Exercises \((2.3E.4)\) and \((2.3E.19)\).)
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Exercise \(\PageIndex{37}\)
\(y''+2y'+10y=4+26x+6x^2+10x^3+e^{x/2}\)
(See Exercises \((2.3E.5)\) and \((2.3E.20)\).)
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Exercise \(\PageIndex{38}\)
\(y''+6y'+10y=22+20x+e^{-3x}\)
(See Exercises \((2.3E.6)\) and \((2.3E.21)\).)
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Exercise \(\PageIndex{39}\)
Prove: If \(y_{p_1}\) is a particular solution of
\begin{eqnarray*}
P_0(x)y''+P_1(x)y'+P_2(x)y=F_1(x)
\end{eqnarray*}
on \((a,b)\) and \(y_{p_2}\) is a particular solution of
\begin{eqnarray*}
P_0(x)y''+P_1(x)y'+P_2(x)y=F_2(x)
\end{eqnarray*}
on \((a,b)\), then \(y_p=y_{p_1}+y_{p_2}\) is a solution of
\begin{eqnarray*}
P_0(x)y''+P_1(x)y'+P_2(x)y=F_1(x)+F_2(x)
\end{eqnarray*}
on \((a,b)\).
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Exercise \(\PageIndex{40}\)
Suppose \(p\), \(q\), and \(f\) are continuous on \((a,b)\). Let \(y_1\), \(y_2\), and \(y_p\) be twice differentiable on \((a,b)\), such that \(y=c_1y_1+c_2y_2+y_p\) is a solution of
\begin{eqnarray*}
y''+p(x)y'+q(x)y=f
\end{eqnarray*}
on \((a,b)\) for every choice of the constants \(c_1,c_2\). Show that \(y_1\) and \(y_2\) are solutions of the complementary equation on \((a,b)\).
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