Internal problem ID [14460]
Internal file name [OUTPUT/14141_Monday_March_25_2024_09_50_25_PM_57340161/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.1, page 141
Problem number: 28.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "reduction_of_order", "second_order_linear_constant_coeff", "linear_second_order_ode_solved_by_an_integrating_factor"
Maple gives the following as the ode type
[[_2nd_order, _missing_x]]
\[ \boxed {y^{\prime \prime }+10 y^{\prime }+25 y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= {\mathrm e}^{-5 t} \end {align*}
Given one basis solution \(y_{1}\left (t \right )\), then the second basis solution is given by \[ y_{2}\left (t \right ) = y_{1} \left (\int \frac {{\mathrm e}^{-\left (\int p d t \right )}}{y_{1}^{2}}d t \right ) \] Where \(p(x)\) is the coefficient of \(y^{\prime }\) when the ode is written in the normal form \[ y^{\prime \prime }+p \left (t \right ) y^{\prime }+q \left (t \right ) y = f \left (t \right ) \] Looking at the ode to solve shows that \[ p \left (t \right ) = 10 \] Therefore \begin{align*} y_{2}\left (t \right ) &= {\mathrm e}^{-5 t} \left (\int {\mathrm e}^{-\left (\int 10d t \right )} {\mathrm e}^{10 t}d t \right ) \\ y_{2}\left (t \right ) &= {\mathrm e}^{-5 t} \int \frac {{\mathrm e}^{-10 t}}{{\mathrm e}^{-10 t}} , dt \\ y_{2}\left (t \right ) &= {\mathrm e}^{-5 t} \left (\int 1d t \right ) \\ y_{2}\left (t \right ) &= t \,{\mathrm e}^{-5 t} \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= c_{1} {\mathrm e}^{-5 t}+c_{2} t \,{\mathrm e}^{-5 t} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-5 t}+c_{2} t \,{\mathrm e}^{-5 t} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-5 t}+c_{2} t \,{\mathrm e}^{-5 t} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+10 y^{\prime }+25 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+10 r +25=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +5\right )^{2}=0 \\ \bullet & {} & \textrm {Root of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =-5 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-5 t} \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (t \right )\hspace {3pt}\textrm {by}\hspace {3pt} t \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=t \,{\mathrm e}^{-5 t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-5 t}+c_{2} t \,{\mathrm e}^{-5 t} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 14
dsolve([diff(y(t),t$2)+10*diff(y(t),t)+25*y(t)=0,exp(-5*t)],singsol=all)
\[ y \left (t \right ) = {\mathrm e}^{-5 t} \left (c_{2} t +c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.014 (sec). Leaf size: 18
DSolve[y''[t]+10*y'[t]+25*y[t]==0,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to e^{-5 t} (c_2 t+c_1) \]