problem 12

Internal problem ID [11785]
Internal ﬁle name [OUTPUT/11795_Thursday_April_11_2024_08_49_44_PM_54708509/index.tex]

Book: Diﬀerential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi. 2004.
Section: Chapter 4, Section 4.3. The method of undetermined coeﬃcients. Exercises page 151
Problem number: 12.
ODE order: 3.
ODE degree: 1.

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

\[ \boxed {4 y^{\prime \prime \prime }-4 y^{\prime \prime }-5 y^{\prime }+3 y=3 x^{3}-8 x} \] 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 \[ 4 y^{\prime \prime \prime }-4 y^{\prime \prime }-5 y^{\prime }+3 y = 0 \] The characteristic equation is \[ 4 \lambda ^{3}-4 \lambda ^{2}-5 \lambda +3 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= {\frac {1}{2}}\\ \lambda _2 &= {\frac {3}{2}}\\ \lambda _3 &= -1 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+{\mathrm e}^{\frac {x}{2}} c_{2} +{\mathrm e}^{\frac {3 x}{2}} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= {\mathrm e}^{\frac {x}{2}} \\ y_3 &= {\mathrm e}^{\frac {3 x}{2}} \\ \end{align*} Now the particular solution to the given ODE is found \[ 4 y^{\prime \prime \prime }-4 y^{\prime \prime }-5 y^{\prime }+3 y = 3 x^{3}-8 x \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ x^{3}+x \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1, x, x^{2}, x^{3}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{{\mathrm e}^{-x}, {\mathrm e}^{\frac {x}{2}}, {\mathrm e}^{\frac {3 x}{2}}\right \} \] Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set. \[ y_p = A_{4} x^{3}+A_{3} x^{2}+A_{2} x +A_{1} \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}\}\) 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 \[ 3 A_{4} x^{3}+3 A_{3} x^{2}-15 x^{2} A_{4}+3 A_{2} x -10 x A_{3}-24 x A_{4}+3 A_{1}-5 A_{2}-8 A_{3}+24 A_{4} = 3 x^{3}-8 x \] Solving for the unknowns by comparing coeﬃcients results in \[ [A_{1} = 42, A_{2} = 22, A_{3} = 5, A_{4} = 1] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = x^{3}+5 x^{2}+22 x +42 \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-x}+{\mathrm e}^{\frac {x}{2}} c_{2} +{\mathrm e}^{\frac {3 x}{2}} c_{3}\right ) + \left (x^{3}+5 x^{2}+22 x +42\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-x}+{\mathrm e}^{\frac {x}{2}} c_{2} +{\mathrm e}^{\frac {3 x}{2}} c_{3} +x^{3}+5 x^{2}+22 x +42 \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{1} {\mathrm e}^{-x}+{\mathrm e}^{\frac {x}{2}} c_{2} +{\mathrm e}^{\frac {3 x}{2}} c_{3} +x^{3}+5 x^{2}+22 x +42 \] Veriﬁed OK.

11.12.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y^{\prime \prime \prime }-4 y^{\prime \prime }-5 y^{\prime }+3 y=3 x^{3}-8 x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=y^{\prime \prime }+\frac {5 y^{\prime }}{4}-\frac {3 y}{4}+\frac {3 x^{3}}{4}-2 x \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }-y^{\prime \prime }-\frac {5 y^{\prime }}{4}+\frac {3 y}{4}=\frac {3}{4} x^{3}-2 x \\ \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 )=y^{\prime \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 )=\frac {3 x^{3}}{4}-2 x +y_{3}\left (x \right )+\frac {5 y_{2}\left (x \right )}{4}-\frac {3 y_{1}\left (x \right )}{4} \\ & {} & \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 )=\frac {3 x^{3}}{4}-2 x +y_{3}\left (x \right )+\frac {5 y_{2}\left (x \right )}{4}-\frac {3 y_{1}\left (x \right )}{4}\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 \\ -\frac {3}{4} & \frac {5}{4} & 1 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ \frac {3}{4} x^{3}-2 x \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ \frac {3}{4} x^{3}-2 x \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -\frac {3}{4} & \frac {5}{4} & 1 \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 [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [\frac {1}{2}, \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]\right ], \left [\frac {3}{2}, \left [\begin {array}{c} \frac {4}{9} \\ \frac {2}{3} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [\frac {1}{2}, \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{\frac {x}{2}}\cdot \left [\begin {array}{c} 4 \\ 2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [\frac {3}{2}, \left [\begin {array}{c} \frac {4}{9} \\ \frac {2}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{\frac {3 x}{2}}\cdot \left [\begin {array}{c} \frac {4}{9} \\ \frac {2}{3} \\ 1 \end {array}\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}+c_{3} {\moverset {\rightarrow }{y}}_{3}+{\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} {\mathrm e}^{-x} & 4 \,{\mathrm e}^{\frac {x}{2}} & \frac {4 \,{\mathrm e}^{\frac {3 x}{2}}}{9} \\ -{\mathrm e}^{-x} & 2 \,{\mathrm e}^{\frac {x}{2}} & \frac {2 \,{\mathrm e}^{\frac {3 x}{2}}}{3} \\ {\mathrm e}^{-x} & {\mathrm e}^{\frac {x}{2}} & {\mathrm e}^{\frac {3 x}{2}} \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}{ccc} {\mathrm e}^{-x} & 4 \,{\mathrm e}^{\frac {x}{2}} & \frac {4 \,{\mathrm e}^{\frac {3 x}{2}}}{9} \\ -{\mathrm e}^{-x} & 2 \,{\mathrm e}^{\frac {x}{2}} & \frac {2 \,{\mathrm e}^{\frac {3 x}{2}}}{3} \\ {\mathrm e}^{-x} & {\mathrm e}^{\frac {x}{2}} & {\mathrm e}^{\frac {3 x}{2}} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & 4 & \frac {4}{9} \\ -1 & 2 & \frac {2}{3} \\ 1 & 1 & 1 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} -\frac {\left ({\mathrm e}^{\frac {5 x}{2}}-5 \,{\mathrm e}^{\frac {3 x}{2}}-1\right ) {\mathrm e}^{-x}}{5} & \frac {\left (3 \,{\mathrm e}^{\frac {5 x}{2}}+5 \,{\mathrm e}^{\frac {3 x}{2}}-8\right ) {\mathrm e}^{-x}}{15} & \frac {2 \left (3 \,{\mathrm e}^{\frac {5 x}{2}}-5 \,{\mathrm e}^{\frac {3 x}{2}}+2\right ) {\mathrm e}^{-x}}{15} \\ -\frac {\left (3 \,{\mathrm e}^{\frac {5 x}{2}}-5 \,{\mathrm e}^{\frac {3 x}{2}}+2\right ) {\mathrm e}^{-x}}{10} & \frac {\left (9 \,{\mathrm e}^{\frac {5 x}{2}}+5 \,{\mathrm e}^{\frac {3 x}{2}}+16\right ) {\mathrm e}^{-x}}{30} & \frac {\left (9 \,{\mathrm e}^{\frac {5 x}{2}}-5 \,{\mathrm e}^{\frac {3 x}{2}}-4\right ) {\mathrm e}^{-x}}{15} \\ -\frac {\left (9 \,{\mathrm e}^{\frac {5 x}{2}}-5 \,{\mathrm e}^{\frac {3 x}{2}}-4\right ) {\mathrm e}^{-x}}{20} & \frac {\left (27 \,{\mathrm e}^{\frac {5 x}{2}}+5 \,{\mathrm e}^{\frac {3 x}{2}}-32\right ) {\mathrm e}^{-x}}{60} & \frac {\left (27 \,{\mathrm e}^{\frac {5 x}{2}}-5 \,{\mathrm e}^{\frac {3 x}{2}}+8\right ) {\mathrm e}^{-x}}{30} \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} {\mathrm e}^{-x} \left (-\frac {128 \,{\mathrm e}^{\frac {3 x}{2}}}{3}+\frac {2}{3}+\left (x^{3}+5 x^{2}+22 x +42\right ) {\mathrm e}^{x}\right ) \\ \frac {\left (9 \,{\mathrm e}^{x} x^{2}-64 \,{\mathrm e}^{\frac {3 x}{2}}+30 x \,{\mathrm e}^{x}+66 \,{\mathrm e}^{x}-2\right ) {\mathrm e}^{-x}}{3} \\ \frac {2 \left (-16 \,{\mathrm e}^{\frac {3 x}{2}}+9 x \,{\mathrm e}^{x}+15 \,{\mathrm e}^{x}+1\right ) {\mathrm e}^{-x}}{3} \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}+c_{3} {\moverset {\rightarrow }{y}}_{3}+\left [\begin {array}{c} {\mathrm e}^{-x} \left (-\frac {128 \,{\mathrm e}^{\frac {3 x}{2}}}{3}+\frac {2}{3}+\left (x^{3}+5 x^{2}+22 x +42\right ) {\mathrm e}^{x}\right ) \\ \frac {\left (9 \,{\mathrm e}^{x} x^{2}-64 \,{\mathrm e}^{\frac {3 x}{2}}+30 x \,{\mathrm e}^{x}+66 \,{\mathrm e}^{x}-2\right ) {\mathrm e}^{-x}}{3} \\ \frac {2 \left (-16 \,{\mathrm e}^{\frac {3 x}{2}}+9 x \,{\mathrm e}^{x}+15 \,{\mathrm e}^{x}+1\right ) {\mathrm e}^{-x}}{3} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-x} \left (\frac {2}{3}+4 \left (-\frac {32}{3}+c_{2} \right ) {\mathrm e}^{\frac {3 x}{2}}+\frac {4 \,{\mathrm e}^{\frac {5 x}{2}} c_{3}}{9}+\left (x^{3}+5 x^{2}+22 x +42\right ) {\mathrm e}^{x}+c_{1} \right ) \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] 
trying high order linear exact nonhomogeneous 
trying differential order: 3; missing the dependent variable 
checking if the LODE has constant coefficients 
<- constant coefficients successful`

\[ y \left (x \right ) = \left (c_{2} {\mathrm e}^{\frac {3 x}{2}}+c_{3} {\mathrm e}^{\frac {5 x}{2}}+\left (x^{3}+5 x^{2}+22 x +42\right ) {\mathrm e}^{x}+c_{1} \right ) {\mathrm e}^{-x} \]

11.12 problem 12

11.12.1 Maple step by step solution