9.19 problem 14.2 (i)

9.19.1 Maple step by step solution

Internal problem ID [13543]
Internal file name [OUTPUT/12715_Friday_February_16_2024_12_11_11_AM_17736975/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 (i).
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "reduction_of_order", "second_order_linear_constant_coeff", "second_order_ode_can_be_made_integrable"

Maple gives the following as the ode type

[[_2nd_order, _missing_x]]

\[ \boxed {y^{\prime \prime }+y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= \sin \left (x \right ) \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 \[ p \left (x \right ) = 0 \] Therefore \begin{align*} y_{2}\left (x \right ) &= \sin \left (x \right ) \left (\int \frac {{\mathrm e}^{-\left (\int 0d x \right )}}{\sin \left (x \right )^{2}}d x \right ) \\ y_{2}\left (x \right ) &= \sin \left (x \right ) \int \frac {1}{\sin \left (x \right )^{2}} , dx \\ y_{2}\left (x \right ) &= \sin \left (x \right ) \left (\int \csc \left (x \right )^{2}d x \right ) \\ y_{2}\left (x \right ) &= -\sin \left (x \right ) \cot \left (x \right ) \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= \sin \left (x \right ) c_{1} -c_{2} \sin \left (x \right ) \cot \left (x \right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \sin \left (x \right ) c_{1} -c_{2} \sin \left (x \right ) \cot \left (x \right ) \\ \end{align*}

Figure 546: Slope field plot

Verification of solutions

\[ y = \sin \left (x \right ) c_{1} -c_{2} \sin \left (x \right ) \cot \left (x \right ) \] Verified OK.

9.19.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+1=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (\mathrm {-I}, \mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )=\cos \left (x \right ) \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=\sin \left (x \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y=\cos \left (x \right ) c_{1} +c_{2} \sin \left (x \right ) \end {array} \]

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: 13

dsolve([diff(y(x),x$2)+y(x)=0,sin(x)],singsol=all)
 

\[ y \left (x \right ) = c_{1} \sin \left (x \right )+c_{2} \cos \left (x \right ) \]

Solution by Mathematica

Time used: 0.013 (sec). Leaf size: 16

DSolve[y''[x]+y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to c_1 \cos (x)+c_2 \sin (x) \]