problem 21.12

Internal problem ID [13683]
Internal ﬁle name [OUTPUT/12855_Saturday_February_17_2024_08_45_31_AM_25215298/index.tex]

Book: Ordinary Diﬀerential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 21. Nonhomogeneous equations in general. Additional exercises page 391
Problem number: 21.12.
ODE order: 4.
ODE degree: 1.

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

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

Looking at the above solution \begin {align*} y = c_{2} x +c_{1} +{\mathrm e}^{i x} c_{3} +{\mathrm e}^{-i x} c_{4} +\frac {x^{2}}{2} \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 \(y = 4\) and \(x = 0\) in the above gives \begin {align*} 4 = 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 x} c_{3} -i {\mathrm e}^{-i x} c_{4} +x \end {align*}

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

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

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

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

substituting \(y^{\prime \prime \prime } = 2\) and \(x = 0\) in the above gives \begin {align*} 2 = -i \left (c_{3} -c_{4} \right )\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}&=3\\ c_{2}&=5\\ c_{3}&=\frac {1}{2}+i\\ c_{4}&=\frac {1}{2}-i \end {align*}

Substituting these values back in above solution results in \begin {align*} y = 5 x +3+\frac {x^{2}}{2}+\cos \left (x \right )-2 \sin \left (x \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 5 x +3+\frac {x^{2}}{2}+\cos \left (x \right )-2 \sin \left (x \right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = 5 x +3+\frac {x^{2}}{2}+\cos \left (x \right )-2 \sin \left (x \right ) \] 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] 
<- differential order: 4; linear nonhomogeneous with symmetry [0,1] successful`

\[ y \left (x \right ) = \frac {x^{2}}{2}+\cos \left (x \right )-2 \sin \left (x \right )+5 x +3 \]

14.10 problem 21.12