Internal problem ID [11497]
Internal file name [OUTPUT/10480_Thursday_May_18_2023_04_20_42_AM_19877591/index.tex]
Book: A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag,
NY. 2015.
Section: Chapter 2, Second order linear equations. Section 2.4.3 Reduction of order. Exercises
page 125
Problem number: 4.
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]]
\[ \boxed {x^{\prime \prime }-2 a x^{\prime }+a^{2} x=0} \] Given that one solution of the ode is \begin {align*} x_1 &= {\mathrm e}^{t a} \end {align*}
Given one basis solution \(x_{1}\left (t \right )\), then the second basis solution is given by \[ x_{2}\left (t \right ) = x_{1} \left (\int \frac {{\mathrm e}^{-\left (\int p d t \right )}}{x_{1}^{2}}d t \right ) \] Where \(p(x)\) is the coefficient of \(x^{\prime }\) when the ode is written in the normal form \[ x^{\prime \prime }+p \left (t \right ) x^{\prime }+q \left (t \right ) x = f \left (t \right ) \] Looking at the ode to solve shows that \[ p \left (t \right ) = -2 a \] Therefore \begin{align*} x_{2}\left (t \right ) &= {\mathrm e}^{t a} \left (\int {\mathrm e}^{-\left (\int -2 a d t \right )} {\mathrm e}^{-2 t a}d t \right ) \\ x_{2}\left (t \right ) &= {\mathrm e}^{t a} \int \frac {{\mathrm e}^{2 t a}}{{\mathrm e}^{2 t a}} , dt \\ x_{2}\left (t \right ) &= {\mathrm e}^{t a} \left (\int 1d t \right ) \\ x_{2}\left (t \right ) &= {\mathrm e}^{t a} t \\ \end{align*} Hence the solution is \begin{align*} x &= c_{1} x_{1}\left (t \right )+c_{2} x_{2}\left (t \right ) \\ &= {\mathrm e}^{t a} c_{1} +c_{2} {\mathrm e}^{t a} t \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} x &= {\mathrm e}^{t a} c_{1} +c_{2} {\mathrm e}^{t a} t \\ \end{align*}
Verification of solutions
\[ x = {\mathrm e}^{t a} c_{1} +c_{2} {\mathrm e}^{t a} t \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{\prime \prime }-2 a x^{\prime }+a^{2} x=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & x^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & a^{2}-2 a r +r^{2}=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (a -r \right )^{2}=0 \\ \bullet & {} & \textrm {Root of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =a \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & x_{1}\left (t \right )={\mathrm e}^{t a} \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} x_{1}\left (t \right )\hspace {3pt}\textrm {by}\hspace {3pt} t \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & x_{2}\left (t \right )={\mathrm e}^{t a} t \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & x=c_{1} x_{1}\left (t \right )+c_{2} x_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & x={\mathrm e}^{t a} c_{1} +c_{2} {\mathrm e}^{t a} t \end {array} \]
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: 0.015 (sec). Leaf size: 14
dsolve([diff(x(t),t$2)-2*a*diff(x(t),t)+a^2*x(t)=0,exp(a*t)],singsol=all)
\[ x \left (t \right ) = {\mathrm e}^{a t} \left (c_{2} t +c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.025 (sec). Leaf size: 18
DSolve[x''[t]-2*a*x'[t]+a^2*x[t]==0,x[t],t,IncludeSingularSolutions -> True]
\[ x(t)\to e^{a t} (c_2 t+c_1) \]