problem 30

Internal problem ID [14704]
Internal ﬁle name [OUTPUT/14384_Wednesday_April_03_2024_02_17_56_PM_58585802/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.6, page 187
Problem number: 30.
ODE order: 4.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime \prime \prime }+y^{\prime \prime }=t} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 1, y^{\prime \prime \prime }\left (0\right ) = 0] \end {align*}

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 \[ y^{\prime \prime \prime \prime }+y^{\prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{4}+\lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 0\\ \lambda _3 &= i\\ \lambda _4 &= -i \end {align*}

Therefore the homogeneous solution is \[ y_h(t)=c_{2} t +c_{1} +{\mathrm e}^{i t} c_{3} +{\mathrm e}^{-i t} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= t \\ y_3 &= {\mathrm e}^{i t} \\ y_4 &= {\mathrm e}^{-i t} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+y^{\prime \prime } = t \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ t \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1, t\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, t, {\mathrm e}^{i t}, {\mathrm e}^{-i t}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes \[ [\{t, t^{2}\}] \] Since \(t\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes \[ [\{t^{2}, t^{3}\}] \] 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. \[ y_p = A_{2} t^{3}+A_{1} t^{2} \] The unknowns \(\{A_{1}, A_{2}\}\) 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 \[ 6 t A_{2}+2 A_{1} = t \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = 0, A_{2} = {\frac {1}{6}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {t^{3}}{6} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{2} t +c_{1} +{\mathrm e}^{i t} c_{3} +{\mathrm e}^{-i t} c_{4}\right ) + \left (\frac {t^{3}}{6}\right ) \\ \end{align*} Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = c_{2} t +c_{1} +{\mathrm e}^{i t} c_{3} +{\mathrm e}^{-i t} c_{4} +\frac {t^{3}}{6} \tag {1} \end {align*}

Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = c_{1} +c_{3} +c_{4}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{2} +i {\mathrm e}^{i t} c_{3} -i {\mathrm e}^{-i t} c_{4} +\frac {t^{2}}{2} \end {align*}

substituting \(y^{\prime } = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = i c_{3} -i c_{4} +c_{2}\tag {2A} \end {align*}

Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = -{\mathrm e}^{i t} c_{3} -{\mathrm e}^{-i t} c_{4} +t \end {align*}

substituting \(y^{\prime \prime } = 1\) and \(t = 0\) in the above gives \begin {align*} 1 = -c_{3} -c_{4}\tag {3A} \end {align*}

Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = -i {\mathrm e}^{i t} c_{3} +i {\mathrm e}^{-i t} c_{4} +1 \end {align*}

substituting \(y^{\prime \prime \prime } = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = -i c_{3} +i c_{4} +1\tag {4A} \end {align*}

Equations {1A,2A,3A,4A} are now solved for \(\{c_{1}, c_{2}, c_{3}, c_{4}\}\). Solving for the constants gives \begin {align*} c_{1}&=1\\ c_{2}&=-1\\ c_{3}&=-\frac {1}{2}-\frac {i}{2}\\ c_{4}&=-\frac {1}{2}+\frac {i}{2} \end {align*}

Substituting these values back in above solution results in \begin {align*} y = -t +1+\frac {t^{3}}{6}-\cos \left (t \right )+\sin \left (t \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -t +1+\frac {t^{3}}{6}-\cos \left (t \right )+\sin \left (t \right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = -t +1+\frac {t^{3}}{6}-\cos \left (t \right )+\sin \left (t \right ) \] Veriﬁed OK.

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] 
-> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = -_b(_a)+_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: 4; linear nonhomogeneous with symmetry [0,1] successful`

\[ y \left (t \right ) = \frac {t^{3}}{6}-\cos \left (t \right )+\sin \left (t \right )-t +1 \]

14.30 problem 30