Internal problem ID [14675]
Internal file name [OUTPUT/14355_Wednesday_April_03_2024_02_17_36_PM_6826807/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: 1.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_linear_constant_coefficients_ODE"
Maple gives the following as the ode type
[[_3rd_order, _missing_y]]
\[ \boxed {y^{\prime \prime \prime }+y^{\prime \prime }={\mathrm e}^{t}} \] 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 \[ y_h(t)=c_{1} {\mathrm e}^{-t}+c_{2} +c_{3} t \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-t} \\ y_2 &= 1 \\ y_3 &= t \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+y^{\prime \prime } = {\mathrm e}^{t} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{t} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{t}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, t, {\mathrm e}^{-t}\} \] 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. \[ y_p = A_{1} {\mathrm e}^{t} \] 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 \[ 2 A_{1} {\mathrm e}^{t} = {\mathrm e}^{t} \] 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 \[ y_p = \frac {{\mathrm e}^{t}}{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-t}+c_{2} +c_{3} t\right ) + \left (\frac {{\mathrm e}^{t}}{2}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-t}+c_{2} +c_{3} t +\frac {{\mathrm e}^{t}}{2} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-t}+c_{2} +c_{3} t +\frac {{\mathrm e}^{t}}{2} \] 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) = -_b(_a)+exp(_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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 19
dsolve(diff(y(t),t$3)+diff(y(t),t$2)=exp(t),y(t), singsol=all)
\[ y \left (t \right ) = \frac {{\mathrm e}^{t}}{2}+{\mathrm e}^{-t} c_{1} +t c_{2} +c_{3} \]
✓ Solution by Mathematica
Time used: 0.093 (sec). Leaf size: 27
DSolve[y'''[t]+y''[t]==Exp[t],y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {e^t}{2}+c_1 e^{-t}+c_3 t+c_2 \]