problem 26

Internal problem ID [14700]
Internal ﬁle name [OUTPUT/14380_Wednesday_April_03_2024_02_17_51_PM_72296140/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: 26.
ODE order: 3.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime \prime }-y^{\prime \prime }=3 t^{2}} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0, y^{\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 \[ y^{\prime \prime \prime }-y^{\prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{3}-\lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 0\\ \lambda _3 &= 0 \end {align*}

Therefore the homogeneous solution is \[ y_h(t)=c_{2} t +c_{1} +c_{3} {\mathrm e}^{t} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= t \\ y_3 &= {\mathrm e}^{t} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-y^{\prime \prime } = 3 t^{2} \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ t^{2} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1, t, t^{2}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, t, {\mathrm e}^{t}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes \[ [\{t, t^{2}, t^{3}\}] \] Since \(t\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes \[ [\{t^{2}, t^{3}, t^{4}\}] \] 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. \[ y_p = A_{3} t^{4}+A_{2} t^{3}+A_{1} t^{2} \] 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 \[ -12 t^{2} A_{3}-6 t A_{2}+24 t A_{3}-2 A_{1}+6 A_{2} = 3 t^{2} \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = -3, A_{2} = -1, A_{3} = -{\frac {1}{4}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {1}{4} t^{4}-t^{3}-3 t^{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{2} t +c_{1} +c_{3} {\mathrm e}^{t}\right ) + \left (-\frac {1}{4} t^{4}-t^{3}-3 t^{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} t +c_{1} +c_{3} {\mathrm e}^{t}-\frac {t^{4}}{4}-t^{3}-3 t^{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 \(y = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = c_{1} +c_{3}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{2} +c_{3} {\mathrm e}^{t}-t^{3}-3 t^{2}-6 t \end {align*}

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

Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = c_{3} {\mathrm e}^{t}-3 t^{2}-6 t -6 \end {align*}

substituting \(y^{\prime \prime } = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = c_{3} -6\tag {3A} \end {align*}

Equations {1A,2A,3A} are now solved for \(\{c_{1}, c_{2}, c_{3}\}\). Solving for the constants gives \begin {align*} c_{1}&=-6\\ c_{2}&=-6\\ c_{3}&=6 \end {align*}

Substituting these values back in above solution results in \begin {align*} y = -6-6 t +6 \,{\mathrm e}^{t}-\frac {t^{4}}{4}-t^{3}-3 t^{2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -6-6 t +6 \,{\mathrm e}^{t}-\frac {t^{4}}{4}-t^{3}-3 t^{2} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -6-6 t +6 \,{\mathrm e}^{t}-\frac {t^{4}}{4}-t^{3}-3 t^{2} \] Veriﬁed OK.

Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
-> Calling odsolve with the ODE`, diff(_b(_a), _a) = 3*_a^2+_b(_a), _b(_a)`   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful 
<- high order exact linear fully integrable successful`

\[ y \left (t \right ) = 6 \,{\mathrm e}^{t}-3 t^{2}-t^{3}-\frac {t^{4}}{4}-6 t -6 \]

14.26 problem 26