Internal problem ID [13536]
Internal file name [OUTPUT/12708_Friday_February_16_2024_12_11_09_AM_69952437/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 14. Higher order equations and the reduction of order method. Additional
exercises page 277
Problem number: 14.2 (b).
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 x} \end {align*}
Given one basis solution \(y_{1}\left (x \right )\), then the second basis solution is given by \[ y_{2}\left (x \right ) = y_{1} \left (\int \frac {{\mathrm e}^{-\left (\int p d x \right )}}{y_{1}^{2}}d x \right ) \] Where \(p(x)\) is the coefficient of \(y^{\prime }\) when the ode is written in the normal form \[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y = f \left (x \right ) \] Looking at the ode to solve shows that \[ p \left (x \right ) = -10 \] Therefore \begin{align*} y_{2}\left (x \right ) &= {\mathrm e}^{5 x} \left (\int {\mathrm e}^{-\left (\int \left (-10\right )d x \right )} {\mathrm e}^{-10 x}d x \right ) \\ y_{2}\left (x \right ) &= {\mathrm e}^{5 x} \int \frac {{\mathrm e}^{10 x}}{{\mathrm e}^{10 x}} , dx \\ y_{2}\left (x \right ) &= {\mathrm e}^{5 x} \left (\int 1d x \right ) \\ y_{2}\left (x \right ) &= {\mathrm e}^{5 x} x \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} {\mathrm e}^{5 x}+c_{2} {\mathrm e}^{5 x} x \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{5 x}+c_{2} {\mathrm e}^{5 x} x \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{5 x}+c_{2} {\mathrm e}^{5 x} x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-10 y^{\prime }+25 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-10 r +25=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -5\right )^{2}=0 \\ \bullet & {} & \textrm {Root of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =5 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )={\mathrm e}^{5 x} \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (x \right )\hspace {3pt}\textrm {by}\hspace {3pt} x \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )={\mathrm e}^{5 x} x \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{5 x}+c_{2} {\mathrm e}^{5 x} x \end {array} \]
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(x),x$2)-10*diff(y(x),x)+25*y(x)=0,exp(5*x)],singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{5 x} \left (c_{2} x +c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.015 (sec). Leaf size: 18
DSolve[y''[x]-10*y'[x]+25*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{5 x} (c_2 x+c_1) \]