problem Problem 20

Internal problem ID [2828]
Internal ﬁle name [OUTPUT/2320_Sunday_June_05_2022_02_59_01_AM_7048863/index.tex]

Book: Diﬀerential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 8, Linear diﬀerential equations of order n. Section 8.10, Chapter review. page 575
Problem number: Problem 20.
ODE order: 3.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_linear_constant_coeﬃcients_ODE"

\[ \boxed {y^{\prime \prime \prime }-6 y^{\prime \prime }+25 y^{\prime }=x^{2}} \] 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 \[ y^{\prime \prime \prime }-6 y^{\prime \prime }+25 y^{\prime } = 0 \] The characteristic equation is \[ \lambda ^{3}-6 \lambda ^{2}+25 \lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 3+4 i\\ \lambda _3 &= 3-4 i \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} +{\mathrm e}^{\left (3-4 i\right ) x} c_{2} +{\mathrm e}^{\left (3+4 i\right ) x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= {\mathrm e}^{\left (3-4 i\right ) x} \\ y_3 &= {\mathrm e}^{\left (3+4 i\right ) x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-6 y^{\prime \prime }+25 y^{\prime } = x^{2} \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ x^{2} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1, x, x^{2}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, {\mathrm e}^{\left (3-4 i\right ) x}, {\mathrm e}^{\left (3+4 i\right ) x}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x, x^{2}, x^{3}\}] \] Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set. \[ y_p = A_{3} x^{3}+A_{2} x^{2}+A_{1} x \] The unknowns \(\{A_{1}, A_{2}, A_{3}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coeﬃcients. Substituting the trial solution into the ODE and simplifying gives \[ 75 x^{2} A_{3}+50 x A_{2}-36 x A_{3}+25 A_{1}-12 A_{2}+6 A_{3} = x^{2} \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = {\frac {22}{15625}}, A_{2} = {\frac {6}{625}}, A_{3} = {\frac {1}{75}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {1}{75} x^{3}+\frac {6}{625} x^{2}+\frac {22}{15625} x \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} +{\mathrm e}^{\left (3-4 i\right ) x} c_{2} +{\mathrm e}^{\left (3+4 i\right ) x} c_{3}\right ) + \left (\frac {1}{75} x^{3}+\frac {6}{625} x^{2}+\frac {22}{15625} x\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} +{\mathrm e}^{\left (3-4 i\right ) x} c_{2} +{\mathrm e}^{\left (3+4 i\right ) x} c_{3} +\frac {x^{3}}{75}+\frac {6 x^{2}}{625}+\frac {22 x}{15625} \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{1} +{\mathrm e}^{\left (3-4 i\right ) x} c_{2} +{\mathrm e}^{\left (3+4 i\right ) x} c_{3} +\frac {x^{3}}{75}+\frac {6 x^{2}}{625}+\frac {22 x}{15625} \] Veriﬁed OK.

12.5.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime \prime }-6 \frac {d}{d x}y^{\prime }+25 y^{\prime }=x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \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 {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=x^{2}+6 y_{3}\left (x \right )-25 y_{2}\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_{3}^{\prime }\left (x \right )=x^{2}+6 y_{3}\left (x \right )-25 y_{2}\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 ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -25 & 6 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ x^{2} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ x^{2} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -25 & 6 \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 [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ], \left [3-4 \,\mathrm {I}, \left [\begin {array}{c} -\frac {7}{625}+\frac {24 \,\mathrm {I}}{625} \\ \frac {3}{25}+\frac {4 \,\mathrm {I}}{25} \\ 1 \end {array}\right ]\right ], \left [3+4 \,\mathrm {I}, \left [\begin {array}{c} -\frac {7}{625}-\frac {24 \,\mathrm {I}}{625} \\ \frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [3-4 \,\mathrm {I}, \left [\begin {array}{c} -\frac {7}{625}+\frac {24 \,\mathrm {I}}{625} \\ \frac {3}{25}+\frac {4 \,\mathrm {I}}{25} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (3-4 \,\mathrm {I}\right ) x}\cdot \left [\begin {array}{c} -\frac {7}{625}+\frac {24 \,\mathrm {I}}{625} \\ \frac {3}{25}+\frac {4 \,\mathrm {I}}{25} \\ 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 \\ {} & {} & {\mathrm e}^{3 x}\cdot \left (\cos \left (4 x \right )-\mathrm {I} \sin \left (4 x \right )\right )\cdot \left [\begin {array}{c} -\frac {7}{625}+\frac {24 \,\mathrm {I}}{625} \\ \frac {3}{25}+\frac {4 \,\mathrm {I}}{25} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{3 x}\cdot \left [\begin {array}{c} \left (-\frac {7}{625}+\frac {24 \,\mathrm {I}}{625}\right ) \left (\cos \left (4 x \right )-\mathrm {I} \sin \left (4 x \right )\right ) \\ \left (\frac {3}{25}+\frac {4 \,\mathrm {I}}{25}\right ) \left (\cos \left (4 x \right )-\mathrm {I} \sin \left (4 x \right )\right ) \\ \cos \left (4 x \right )-\mathrm {I} \sin \left (4 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{3 x}\cdot \left [\begin {array}{c} -\frac {7 \cos \left (4 x \right )}{625}+\frac {24 \sin \left (4 x \right )}{625} \\ \frac {3 \cos \left (4 x \right )}{25}+\frac {4 \sin \left (4 x \right )}{25} \\ \cos \left (4 x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{3 x}\cdot \left [\begin {array}{c} \frac {7 \sin \left (4 x \right )}{625}+\frac {24 \cos \left (4 x \right )}{625} \\ -\frac {3 \sin \left (4 x \right )}{25}+\frac {4 \cos \left (4 x \right )}{25} \\ -\sin \left (4 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}+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\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}{ccc} 1 & {\mathrm e}^{3 x} \left (-\frac {7 \cos \left (4 x \right )}{625}+\frac {24 \sin \left (4 x \right )}{625}\right ) & {\mathrm e}^{3 x} \left (\frac {7 \sin \left (4 x \right )}{625}+\frac {24 \cos \left (4 x \right )}{625}\right ) \\ 0 & {\mathrm e}^{3 x} \left (\frac {3 \cos \left (4 x \right )}{25}+\frac {4 \sin \left (4 x \right )}{25}\right ) & {\mathrm e}^{3 x} \left (-\frac {3 \sin \left (4 x \right )}{25}+\frac {4 \cos \left (4 x \right )}{25}\right ) \\ 0 & {\mathrm e}^{3 x} \cos \left (4 x \right ) & -{\mathrm e}^{3 x} \sin \left (4 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 \phi \left (0\right )^{-1} \\ {} & \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}{ccc} 1 & {\mathrm e}^{3 x} \left (-\frac {7 \cos \left (4 x \right )}{625}+\frac {24 \sin \left (4 x \right )}{625}\right ) & {\mathrm e}^{3 x} \left (\frac {7 \sin \left (4 x \right )}{625}+\frac {24 \cos \left (4 x \right )}{625}\right ) \\ 0 & {\mathrm e}^{3 x} \left (\frac {3 \cos \left (4 x \right )}{25}+\frac {4 \sin \left (4 x \right )}{25}\right ) & {\mathrm e}^{3 x} \left (-\frac {3 \sin \left (4 x \right )}{25}+\frac {4 \cos \left (4 x \right )}{25}\right ) \\ 0 & {\mathrm e}^{3 x} \cos \left (4 x \right ) & -{\mathrm e}^{3 x} \sin \left (4 x \right ) \end {array}\right ]\cdot \left [\begin {array}{ccc} 1 & -\frac {7}{625} & \frac {24}{625} \\ 0 & \frac {3}{25} & \frac {4}{25} \\ 0 & 1 & 0 \end {array}\right ]^{-1} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} 1 & -\frac {6}{25}+\frac {{\mathrm e}^{3 x} \left (7 \sin \left (4 x \right )+24 \cos \left (4 x \right )\right )}{100} & \frac {1}{25}+\frac {\left (3 \sin \left (4 x \right )-4 \cos \left (4 x \right )\right ) {\mathrm e}^{3 x}}{100} \\ 0 & \frac {{\mathrm e}^{3 x} \left (-3 \sin \left (4 x \right )+4 \cos \left (4 x \right )\right )}{4} & \frac {{\mathrm e}^{3 x} \sin \left (4 x \right )}{4} \\ 0 & -\frac {25 \,{\mathrm e}^{3 x} \sin \left (4 x \right )}{4} & \frac {{\mathrm e}^{3 x} \left (4 \cos \left (4 x \right )+3 \sin \left (4 x \right )\right )}{4} \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 )=\Phi \left (x \right )^{-1}\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}\Phi \left (s \right )^{-1}\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}\Phi \left (s \right )^{-1}\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 {168}{390625}+\frac {\left (336 \cos \left (4 x \right )-527 \sin \left (4 x \right )\right ) {\mathrm e}^{3 x}}{781250}+\frac {x^{3}}{75}+\frac {6 x^{2}}{625}+\frac {22 x}{15625} \\ \frac {\left (-44 \cos \left (4 x \right )-117 \sin \left (4 x \right )\right ) {\mathrm e}^{3 x}}{31250}+\frac {x^{2}}{25}+\frac {12 x}{625}+\frac {22}{15625} \\ \frac {\left (-7 \sin \left (4 x \right )-24 \cos \left (4 x \right )\right ) {\mathrm e}^{3 x}}{1250}+\frac {2 x}{25}+\frac {12}{625} \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}+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+\left [\begin {array}{c} -\frac {168}{390625}+\frac {\left (336 \cos \left (4 x \right )-527 \sin \left (4 x \right )\right ) {\mathrm e}^{3 x}}{781250}+\frac {x^{3}}{75}+\frac {6 x^{2}}{625}+\frac {22 x}{15625} \\ \frac {\left (-44 \cos \left (4 x \right )-117 \sin \left (4 x \right )\right ) {\mathrm e}^{3 x}}{31250}+\frac {x^{2}}{25}+\frac {12 x}{625}+\frac {22}{15625} \\ \frac {\left (-7 \sin \left (4 x \right )-24 \cos \left (4 x \right )\right ) {\mathrm e}^{3 x}}{1250}+\frac {2 x}{25}+\frac {12}{625} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-\frac {168}{390625}+\frac {\left (\left (168-4375 c_{2} +15000 c_{3} \right ) \cos \left (4 x \right )+15000 \left (c_{2} +\frac {7 c_{3}}{24}-\frac {527}{30000}\right ) \sin \left (4 x \right )\right ) {\mathrm e}^{3 x}}{390625}+\frac {x^{3}}{75}+\frac {6 x^{2}}{625}+\frac {22 x}{15625}+c_{1} \end {array} \]

Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 3; linear nonhomogeneous with symmetry [0,1] 
-> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = _a^2+6*(diff(_b(_a), _a))-25*_b(_a), _b(_a)`   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying high order exact linear fully integrable 
   trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
   trying a double symmetry of the form [xi=0, eta=F(x)] 
   -> Try solving first the homogeneous part of the ODE 
      checking if the LODE has constant coefficients 
      <- constant coefficients successful 
   <- solving first the homogeneous part of the ODE successful 
<- differential order: 3; linear nonhomogeneous with symmetry [0,1] successful`

\[ y \left (x \right ) = \frac {\left (\left (3 c_{1} -4 c_{2} \right ) \cos \left (4 x \right )+4 \sin \left (4 x \right ) \left (c_{1} +\frac {3 c_{2}}{4}\right )\right ) {\mathrm e}^{3 x}}{25}+\frac {x^{3}}{75}+\frac {6 x^{2}}{625}+\frac {22 x}{15625}+c_{3} \]

\[ y(x)\to \frac {x^3}{75}+\frac {6 x^2}{625}+\frac {22 x}{15625}-\frac {1}{25} (4 c_1-3 c_2) e^{3 x} \cos (4 x)+\frac {1}{25} (3 c_1+4 c_2) e^{3 x} \sin (4 x)+c_3 \]

12.5 problem Problem 20

12.5.1 Maple step by step solution