Internal problem ID [3263]
Internal file name [OUTPUT/2755_Sunday_June_05_2022_08_40_05_AM_87500085/index.tex]
Book: Differential equations for engineers by Wei-Chau XIE, Cambridge Press 2010
Section: Chapter 4. Linear Differential Equations. Page 183
Problem number: 22.
ODE order: 4.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_linear_constant_coefficients_ODE"
Maple gives the following as the ode type
[[_high_order, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime \prime \prime }+5 y^{\prime \prime }+4 y=\sin \left (x \right ) \cos \left (2 x \right )} \] This is higher order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE And \(y_p\) is a particular solution to the nonhomogeneous ODE. \(y_h\) is the solution to \[ y^{\prime \prime \prime \prime }+5 y^{\prime \prime }+4 y = 0 \] The characteristic equation is \[ \lambda ^{4}+5 \lambda ^{2}+4 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2 i\\ \lambda _2 &= -2 i\\ \lambda _3 &= i\\ \lambda _4 &= -i \end {align*}
Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-2 i x} c_{1} +{\mathrm e}^{-i x} c_{2} +{\mathrm e}^{i x} c_{3} +{\mathrm e}^{2 i x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-2 i x} \\ y_2 &= {\mathrm e}^{-i x} \\ y_3 &= {\mathrm e}^{i x} \\ y_4 &= {\mathrm e}^{2 i x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+5 y^{\prime \prime }+4 y = \sin \left (x \right ) \cos \left (2 x \right ) \] Let the particular solution be \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3+U_4 y_4 \] Where \(y_i\) are the basis solutions found above for the homogeneous solution \(y_h\) and \(U_i(x)\) are functions to be determined as follows \[ U_i = (-1)^{n-i} \int { \frac {F(x) W_i(x) }{a W(x)} \, dx} \] Where \(W(x)\) is the Wronskian and \(W_i(x)\) is the Wronskian that results after deleting the last row and the \(i\)-th column of the determinant and \(n\) is the order of the ODE or equivalently, the number of basis solutions, and \(a\) is the coefficient of the leading derivative in the ODE, and \(F(x)\) is the RHS of the ODE. Therefore, the first step is to find the Wronskian \(W \left (x \right )\). This is given by \begin {equation*} W(x) = \begin {vmatrix} y_1&y_2&y_3&y_4\\ y_1'&y_2'&y_3'&y_4'\\ y_1''&y_2''&y_3''&y_4''\\ y_1'''&y_2'''&y_3'''&y_4'''\\ \end {vmatrix} \end {equation*} Substituting the fundamental set of solutions \(y_i\) found above in the Wronskian gives \begin {align*} W &= \left [\begin {array}{cccc} {\mathrm e}^{-2 i x} & {\mathrm e}^{-i x} & {\mathrm e}^{i x} & {\mathrm e}^{2 i x} \\ -2 i {\mathrm e}^{-2 i x} & -i {\mathrm e}^{-i x} & i {\mathrm e}^{i x} & 2 i {\mathrm e}^{2 i x} \\ -4 \,{\mathrm e}^{-2 i x} & -{\mathrm e}^{-i x} & -{\mathrm e}^{i x} & -4 \,{\mathrm e}^{2 i x} \\ 8 i {\mathrm e}^{-2 i x} & i {\mathrm e}^{-i x} & -i {\mathrm e}^{i x} & -8 i {\mathrm e}^{2 i x} \end {array}\right ] \\ |W| &= -72 \,{\mathrm e}^{-2 i x} {\mathrm e}^{-i x} {\mathrm e}^{2 i x} {\mathrm e}^{i x} \end {align*}
The determinant simplifies to \begin {align*} |W| &= -72 \end {align*}
Now we determine \(W_i\) for each \(U_i\). \begin {align*} W_1(x) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{-i x} & {\mathrm e}^{i x} & {\mathrm e}^{2 i x} \\ -i {\mathrm e}^{-i x} & i {\mathrm e}^{i x} & 2 i {\mathrm e}^{2 i x} \\ -{\mathrm e}^{-i x} & -{\mathrm e}^{i x} & -4 \,{\mathrm e}^{2 i x} \end {array}\right ] \\ &= -6 i {\mathrm e}^{2 i x} \end {align*}
\begin {align*} W_2(x) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{-2 i x} & {\mathrm e}^{i x} & {\mathrm e}^{2 i x} \\ -2 i {\mathrm e}^{-2 i x} & i {\mathrm e}^{i x} & 2 i {\mathrm e}^{2 i x} \\ -4 \,{\mathrm e}^{-2 i x} & -{\mathrm e}^{i x} & -4 \,{\mathrm e}^{2 i x} \end {array}\right ] \\ &= -12 i {\mathrm e}^{i x} \end {align*}
\begin {align*} W_3(x) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{-2 i x} & {\mathrm e}^{-i x} & {\mathrm e}^{2 i x} \\ -2 i {\mathrm e}^{-2 i x} & -i {\mathrm e}^{-i x} & 2 i {\mathrm e}^{2 i x} \\ -4 \,{\mathrm e}^{-2 i x} & -{\mathrm e}^{-i x} & -4 \,{\mathrm e}^{2 i x} \end {array}\right ] \\ &= -12 i {\mathrm e}^{-i x} \end {align*}
\begin {align*} W_4(x) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{-2 i x} & {\mathrm e}^{-i x} & {\mathrm e}^{i x} \\ -2 i {\mathrm e}^{-2 i x} & -i {\mathrm e}^{-i x} & i {\mathrm e}^{i x} \\ -4 \,{\mathrm e}^{-2 i x} & -{\mathrm e}^{-i x} & -{\mathrm e}^{i x} \end {array}\right ] \\ &= -6 i {\mathrm e}^{-2 i x} \end {align*}
Now we are ready to evaluate each \(U_i(x)\). \begin {align*} U_1 &= (-1)^{4-1} \int { \frac {F(x) W_1(x) }{a W(x)} \, dx}\\ &= (-1)^{3} \int { \frac { \left (\sin \left (x \right ) \cos \left (2 x \right )\right ) \left (-6 i {\mathrm e}^{2 i x}\right )}{\left (1\right ) \left (-72\right )} \, dx} \\ &= - \int { \frac {-6 i \sin \left (x \right ) \cos \left (2 x \right ) {\mathrm e}^{2 i x}}{-72} \, dx}\\ &= - \int {\left (\frac {i \sin \left (x \right ) \cos \left (2 x \right ) {\mathrm e}^{2 i x}}{12}\right ) \, dx}\\ &= -\frac {i \left (-\frac {{\mathrm e}^{2 i x} \cos \left (x \right )}{6}+\frac {i {\mathrm e}^{2 i x} \sin \left (x \right )}{3}-\frac {3 \,{\mathrm e}^{2 i x} \cos \left (3 x \right )}{10}+\frac {i {\mathrm e}^{2 i x} \sin \left (3 x \right )}{5}\right )}{12} \end {align*}
\begin {align*} U_2 &= (-1)^{4-2} \int { \frac {F(x) W_2(x) }{a W(x)} \, dx}\\ &= (-1)^{2} \int { \frac { \left (\sin \left (x \right ) \cos \left (2 x \right )\right ) \left (-12 i {\mathrm e}^{i x}\right )}{\left (1\right ) \left (-72\right )} \, dx} \\ &= \int { \frac {-12 i \sin \left (x \right ) \cos \left (2 x \right ) {\mathrm e}^{i x}}{-72} \, dx}\\ &= \int {\left (\frac {i \sin \left (x \right ) \cos \left (2 x \right ) {\mathrm e}^{i x}}{6}\right ) \, dx}\\ &= \int \frac {i \sin \left (x \right ) \cos \left (2 x \right ) {\mathrm e}^{i x}}{6}d x \end {align*}
\begin {align*} U_3 &= (-1)^{4-3} \int { \frac {F(x) W_3(x) }{a W(x)} \, dx}\\ &= (-1)^{1} \int { \frac { \left (\sin \left (x \right ) \cos \left (2 x \right )\right ) \left (-12 i {\mathrm e}^{-i x}\right )}{\left (1\right ) \left (-72\right )} \, dx} \\ &= - \int { \frac {-12 i \sin \left (x \right ) \cos \left (2 x \right ) {\mathrm e}^{-i x}}{-72} \, dx}\\ &= - \int {\left (\frac {i \sin \left (x \right ) \cos \left (2 x \right ) {\mathrm e}^{-i x}}{6}\right ) \, dx}\\ &= -\left (\int \frac {i \sin \left (x \right ) \cos \left (2 x \right ) {\mathrm e}^{-i x}}{6}d x \right ) \end {align*}
\begin {align*} U_4 &= (-1)^{4-4} \int { \frac {F(x) W_4(x) }{a W(x)} \, dx}\\ &= (-1)^{0} \int { \frac { \left (\sin \left (x \right ) \cos \left (2 x \right )\right ) \left (-6 i {\mathrm e}^{-2 i x}\right )}{\left (1\right ) \left (-72\right )} \, dx} \\ &= \int { \frac {-6 i \sin \left (x \right ) \cos \left (2 x \right ) {\mathrm e}^{-2 i x}}{-72} \, dx}\\ &= \int {\left (\frac {i \sin \left (x \right ) \cos \left (2 x \right ) {\mathrm e}^{-2 i x}}{12}\right ) \, dx}\\ &= \frac {i \left (-\frac {{\mathrm e}^{-2 i x} \cos \left (x \right )}{6}-\frac {i {\mathrm e}^{-2 i x} \sin \left (x \right )}{3}-\frac {3 \,{\mathrm e}^{-2 i x} \cos \left (3 x \right )}{10}-\frac {i {\mathrm e}^{-2 i x} \sin \left (3 x \right )}{5}\right )}{12} \end {align*}
Now that all the \(U_i\) functions have been determined, the particular solution is found from \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3+U_4 y_4 \] Hence \begin {equation*} \begin {split} y_p &= \left (-\frac {i \left (-\frac {{\mathrm e}^{2 i x} \cos \left (x \right )}{6}+\frac {i {\mathrm e}^{2 i x} \sin \left (x \right )}{3}-\frac {3 \,{\mathrm e}^{2 i x} \cos \left (3 x \right )}{10}+\frac {i {\mathrm e}^{2 i x} \sin \left (3 x \right )}{5}\right )}{12}\right ) \left ({\mathrm e}^{-2 i x}\right ) \\ &+\left (\int \frac {i \sin \left (x \right ) \cos \left (2 x \right ) {\mathrm e}^{i x}}{6}d x\right ) \left ({\mathrm e}^{-i x}\right ) \\ &+\left (-\left (\int \frac {i \sin \left (x \right ) \cos \left (2 x \right ) {\mathrm e}^{-i x}}{6}d x \right )\right ) \left ({\mathrm e}^{i x}\right ) \\ &+\left (\frac {i \left (-\frac {{\mathrm e}^{-2 i x} \cos \left (x \right )}{6}-\frac {i {\mathrm e}^{-2 i x} \sin \left (x \right )}{3}-\frac {3 \,{\mathrm e}^{-2 i x} \cos \left (3 x \right )}{10}-\frac {i {\mathrm e}^{-2 i x} \sin \left (3 x \right )}{5}\right )}{12}\right ) \left ({\mathrm e}^{2 i x}\right ) \end {split} \end {equation*} Therefore the particular solution is \[ y_p = \frac {2 \sin \left (x \right ) \cos \left (x \right )^{2}}{15}+\frac {\sin \left (x \right )}{45}-\frac {i \left (\int \sin \left (x \right ) \cos \left (2 x \right ) {\mathrm e}^{-i x}d x \right ) {\mathrm e}^{i x}}{6}+\frac {i \left (\int \sin \left (x \right ) \cos \left (2 x \right ) {\mathrm e}^{i x}d x \right ) {\mathrm e}^{-i x}}{6} \] Which simplifies to \[ y_p = -\frac {\left (\int \cos \left (2 x \right ) \sin \left (x \right )^{2}d x \right ) \cos \left (x \right )}{3}+\frac {2 \sin \left (x \right ) \left (\cos \left (x \right )^{2}+\frac {5 \left (\int \sin \left (4 x \right )d x \right )}{8}+\frac {1}{6}\right )}{15} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-2 i x} c_{1} +{\mathrm e}^{-i x} c_{2} +{\mathrm e}^{i x} c_{3} +{\mathrm e}^{2 i x} c_{4}\right ) + \left (-\frac {\left (\int \cos \left (2 x \right ) \sin \left (x \right )^{2}d x \right ) \cos \left (x \right )}{3}+\frac {2 \sin \left (x \right ) \left (\cos \left (x \right )^{2}+\frac {5 \left (\int \sin \left (4 x \right )d x \right )}{8}+\frac {1}{6}\right )}{15}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-2 i x} c_{1} +{\mathrm e}^{-i x} c_{2} +{\mathrm e}^{i x} c_{3} +{\mathrm e}^{2 i x} c_{4} -\frac {\left (\int \cos \left (2 x \right ) \sin \left (x \right )^{2}d x \right ) \cos \left (x \right )}{3}+\frac {2 \sin \left (x \right ) \left (\cos \left (x \right )^{2}+\frac {5 \left (\int \sin \left (4 x \right )d x \right )}{8}+\frac {1}{6}\right )}{15} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{-2 i x} c_{1} +{\mathrm e}^{-i x} c_{2} +{\mathrm e}^{i x} c_{3} +{\mathrm e}^{2 i x} c_{4} -\frac {\left (\int \cos \left (2 x \right ) \sin \left (x \right )^{2}d x \right ) \cos \left (x \right )}{3}+\frac {2 \sin \left (x \right ) \left (\cos \left (x \right )^{2}+\frac {5 \left (\int \sin \left (4 x \right )d x \right )}{8}+\frac {1}{6}\right )}{15} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\prime \prime }+5 \frac {d}{d x}y^{\prime }+4 y=\sin \left (x \right ) \cos \left (2 x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\prime \prime } \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (x \right ) \\ {} & {} & y_{1}\left (x \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=\frac {d}{d x}y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{4}\left (x \right ) \\ {} & {} & y_{4}\left (x \right )=\frac {d}{d x}y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{4}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{4}^{\prime }\left (x \right )=\sin \left (x \right ) \cos \left (2 x \right )-5 y_{3}\left (x \right )-4 y_{1}\left (x \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=y_{1}^{\prime }\left (x \right ), y_{3}\left (x \right )=y_{2}^{\prime }\left (x \right ), y_{4}\left (x \right )=y_{3}^{\prime }\left (x \right ), y_{4}^{\prime }\left (x \right )=\sin \left (x \right ) \cos \left (2 x \right )-5 y_{3}\left (x \right )-4 y_{1}\left (x \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{c} y_{1}\left (x \right ) \\ y_{2}\left (x \right ) \\ y_{3}\left (x \right ) \\ y_{4}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -4 & 0 & -5 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ \sin \left (x \right ) \cos \left (2 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ \sin \left (x \right ) \cos \left (2 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -4 & 0 & -5 & 0 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=A \cdot {\moverset {\rightarrow }{y}}\left (x \right )+{\moverset {\rightarrow }{f}} \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [\mathrm {-I}, \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ]\right ], \left [\mathrm {I}, \left [\begin {array}{c} \mathrm {I} \\ -1 \\ \mathrm {-I} \\ 1 \end {array}\right ]\right ], \left [2 \,\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ -\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 \,\mathrm {I} x}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Use Euler identity to write solution in terms of}\hspace {3pt} \sin \hspace {3pt}\textrm {and}\hspace {3pt} \cos \\ {} & {} & \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right )\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ -\frac {\cos \left (2 x \right )}{4}+\frac {\mathrm {I} \sin \left (2 x \right )}{4} \\ \frac {\mathrm {I}}{2} \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ \cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{1}\left (x \right )=\left [\begin {array}{c} -\frac {\sin \left (2 x \right )}{8} \\ -\frac {\cos \left (2 x \right )}{4} \\ \frac {\sin \left (2 x \right )}{2} \\ \cos \left (2 x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{2}\left (x \right )=\left [\begin {array}{c} -\frac {\cos \left (2 x \right )}{8} \\ \frac {\sin \left (2 x \right )}{4} \\ \frac {\cos \left (2 x \right )}{2} \\ -\sin \left (2 x \right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [\mathrm {-I}, \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\mathrm {-I} x}\cdot \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Use Euler identity to write solution in terms of}\hspace {3pt} \sin \hspace {3pt}\textrm {and}\hspace {3pt} \cos \\ {} & {} & \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right )\cdot \left [\begin {array}{c} \mathrm {-I} \\ -1 \\ \mathrm {I} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} \mathrm {-I} \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ -\cos \left (x \right )+\mathrm {I} \sin \left (x \right ) \\ \mathrm {I} \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ \cos \left (x \right )-\mathrm {I} \sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{3}\left (x \right )=\left [\begin {array}{c} -\sin \left (x \right ) \\ -\cos \left (x \right ) \\ \sin \left (x \right ) \\ \cos \left (x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (x \right )=\left [\begin {array}{c} -\cos \left (x \right ) \\ \sin \left (x \right ) \\ \cos \left (x \right ) \\ -\sin \left (x \right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (x \right )+{\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (x \right )=\left [\begin {array}{cccc} -\frac {\sin \left (2 x \right )}{8} & -\frac {\cos \left (2 x \right )}{8} & -\sin \left (x \right ) & -\cos \left (x \right ) \\ -\frac {\cos \left (2 x \right )}{4} & \frac {\sin \left (2 x \right )}{4} & -\cos \left (x \right ) & \sin \left (x \right ) \\ \frac {\sin \left (2 x \right )}{2} & \frac {\cos \left (2 x \right )}{2} & \sin \left (x \right ) & \cos \left (x \right ) \\ \cos \left (2 x \right ) & -\sin \left (2 x \right ) & \cos \left (x \right ) & -\sin \left (x \right ) \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (x \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {satisfying}\hspace {3pt} \Phi \left (0\right )=I \hspace {3pt}\textrm {where}\hspace {3pt} I \hspace {3pt}\textrm {is the identity matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\phi \left (x \right )\cdot \frac {1}{\phi \left (0\right )} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{cccc} -\frac {\sin \left (2 x \right )}{8} & -\frac {\cos \left (2 x \right )}{8} & -\sin \left (x \right ) & -\cos \left (x \right ) \\ -\frac {\cos \left (2 x \right )}{4} & \frac {\sin \left (2 x \right )}{4} & -\cos \left (x \right ) & \sin \left (x \right ) \\ \frac {\sin \left (2 x \right )}{2} & \frac {\cos \left (2 x \right )}{2} & \sin \left (x \right ) & \cos \left (x \right ) \\ \cos \left (2 x \right ) & -\sin \left (2 x \right ) & \cos \left (x \right ) & -\sin \left (x \right ) \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{cccc} 0 & -\frac {1}{8} & 0 & -1 \\ -\frac {1}{4} & 0 & -1 & 0 \\ 0 & \frac {1}{2} & 0 & 1 \\ 1 & 0 & 1 & 0 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{cccc} -\frac {2 \cos \left (x \right )^{2}}{3}+\frac {1}{3}+\frac {4 \cos \left (x \right )}{3} & -\frac {\sin \left (2 x \right )}{6}+\frac {4 \sin \left (x \right )}{3} & -\frac {2 \cos \left (x \right )^{2}}{3}+\frac {1}{3}+\frac {\cos \left (x \right )}{3} & -\frac {\sin \left (2 x \right )}{6}+\frac {\sin \left (x \right )}{3} \\ \frac {2 \sin \left (2 x \right )}{3}-\frac {4 \sin \left (x \right )}{3} & -\frac {2 \cos \left (x \right )^{2}}{3}+\frac {1}{3}+\frac {4 \cos \left (x \right )}{3} & \frac {2 \sin \left (2 x \right )}{3}-\frac {\sin \left (x \right )}{3} & -\frac {2 \cos \left (x \right )^{2}}{3}+\frac {1}{3}+\frac {\cos \left (x \right )}{3} \\ \frac {8 \cos \left (x \right )^{2}}{3}-\frac {4}{3}-\frac {4 \cos \left (x \right )}{3} & \frac {2 \sin \left (2 x \right )}{3}-\frac {4 \sin \left (x \right )}{3} & \frac {8 \cos \left (x \right )^{2}}{3}-\frac {4}{3}-\frac {\cos \left (x \right )}{3} & \frac {2 \sin \left (2 x \right )}{3}-\frac {\sin \left (x \right )}{3} \\ -\frac {8 \sin \left (2 x \right )}{3}+\frac {4 \sin \left (x \right )}{3} & \frac {8 \cos \left (x \right )^{2}}{3}-\frac {4}{3}-\frac {4 \cos \left (x \right )}{3} & -\frac {8 \sin \left (2 x \right )}{3}+\frac {\sin \left (x \right )}{3} & \frac {8 \cos \left (x \right )^{2}}{3}-\frac {4}{3}-\frac {\cos \left (x \right )}{3} \end {array}\right ] \\ \square & {} & \textrm {Find a particular solution of the system of ODEs using variation of parameters}\hspace {3pt} \\ {} & \circ & \textrm {Let the particular solution be the fundamental matrix multiplied by}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (x \right )=\Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {The fundamental matrix has columns that are solutions to the homogeneous system so its derivative follows that of the homogeneous system}\hspace {3pt} \\ {} & {} & A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=\frac {1}{\Phi \left (x \right )}\cdot {\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (x \right )=\int _{0}^{x}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot \left (\int _{0}^{x}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \right ) \\ {} & \circ & \textrm {Plug in the fundamental matrix and the forcing function and compute}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\left [\begin {array}{c} \frac {\sin \left (x \right ) \cos \left (x \right )^{2}}{20}+\frac {\left (15 x -28 \sin \left (x \right )\right ) \cos \left (x \right )}{180}+\frac {\sin \left (x \right )}{45} \\ \frac {3 \cos \left (x \right )^{3}}{20}-\frac {14 \cos \left (x \right )^{2}}{45}-\frac {x \sin \left (x \right )}{12}+\frac {\cos \left (x \right )}{180}+\frac {7}{45} \\ -\frac {9 \sin \left (x \right ) \cos \left (x \right )^{2}}{20}+\frac {\left (-15 x +112 \sin \left (x \right )\right ) \cos \left (x \right )}{180}-\frac {4 \sin \left (x \right )}{45} \\ -\frac {27 \cos \left (x \right )^{3}}{20}+\frac {56 \cos \left (x \right )^{2}}{45}+\frac {x \sin \left (x \right )}{12}+\frac {131 \cos \left (x \right )}{180}-\frac {28}{45} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (x \right )+\left [\begin {array}{c} \frac {\sin \left (x \right ) \cos \left (x \right )^{2}}{20}+\frac {\left (15 x -28 \sin \left (x \right )\right ) \cos \left (x \right )}{180}+\frac {\sin \left (x \right )}{45} \\ \frac {3 \cos \left (x \right )^{3}}{20}-\frac {14 \cos \left (x \right )^{2}}{45}-\frac {x \sin \left (x \right )}{12}+\frac {\cos \left (x \right )}{180}+\frac {7}{45} \\ -\frac {9 \sin \left (x \right ) \cos \left (x \right )^{2}}{20}+\frac {\left (-15 x +112 \sin \left (x \right )\right ) \cos \left (x \right )}{180}-\frac {4 \sin \left (x \right )}{45} \\ -\frac {27 \cos \left (x \right )^{3}}{20}+\frac {56 \cos \left (x \right )^{2}}{45}+\frac {x \sin \left (x \right )}{12}+\frac {131 \cos \left (x \right )}{180}-\frac {28}{45} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (-5 c_{2} +\sin \left (x \right )\right ) \cos \left (x \right )^{2}}{20}+\frac {\left (\left (-45 c_{1} -28\right ) \sin \left (x \right )+15 x -180 c_{4} \right ) \cos \left (x \right )}{180}+\frac {\left (1-45 c_{3} \right ) \sin \left (x \right )}{45}+\frac {c_{2}}{8} \end {array} \]
Maple trace
`Methods for high order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 4; linear nonhomogeneous with symmetry [0,1] trying high order linear exact nonhomogeneous trying differential order: 4; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 43
dsolve(diff(y(x),x$4)+5*diff(y(x),x$2)+4*y(x)=sin(x)*cos(2*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (40 c_{3} +\sin \left (x \right )\right ) \cos \left (x \right )^{2}}{20}+\frac {\left (24 c_{4} \sin \left (x \right )+x +12 c_{1} \right ) \cos \left (x \right )}{12}+\frac {\left (360 c_{2} -7\right ) \sin \left (x \right )}{360}-c_{3} \]
✓ Solution by Mathematica
Time used: 0.04 (sec). Leaf size: 50
DSolve[y''''[x]+5*y''[x]+4*y[x]==Sin[x]*Cos[2*x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {\sin (x)}{72}+\frac {1}{80} \sin (3 x)+\left (\frac {x}{12}+c_3\right ) \cos (x)+c_1 \cos (2 x)+c_4 \sin (x)+c_2 \sin (2 x) \]