Internal problem ID [13547]
Internal file name [OUTPUT/12719_Friday_February_16_2024_12_11_13_AM_11325384/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 (m).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "reduction_of_order", "second_order_euler_ode", "second_order_change_of_variable_on_x_method_1", "second_order_change_of_variable_on_x_method_2", "second_order_change_of_variable_on_y_method_2"
Maple gives the following as the ode type
[[_Emden, _Fowler], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]
\[ \boxed {x^{2} y^{\prime \prime }+y^{\prime } x +y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= \frac {x}{2}-\frac {1}{2 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 \[ p \left (x \right ) = \frac {1}{x} \] Therefore \begin{align*} y_{2}\left (x \right ) &= \left (\frac {x}{2}-\frac {1}{2 x}\right ) \left (\int \frac {{\mathrm e}^{-\left (\int \frac {1}{x}d x \right )}}{\left (\frac {x}{2}-\frac {1}{2 x}\right )^{2}}d x \right ) \\ y_{2}\left (x \right ) &= \frac {x}{2}-\frac {1}{2 x} \int \frac {\frac {1}{x}}{\left (\frac {x}{2}-\frac {1}{2 x}\right )^{2}} , dx \\ y_{2}\left (x \right ) &= \left (\frac {x}{2}-\frac {1}{2 x}\right ) \left (\int \frac {4 x}{\left (x^{2}-1\right )^{2}}d x \right ) \\ y_{2}\left (x \right ) &= -\frac {2 \left (\frac {x}{2}-\frac {1}{2 x}\right )}{x^{2}-1} \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= \left (\frac {x}{2}-\frac {1}{2 x}\right ) c_{1} -\frac {2 c_{2} \left (\frac {x}{2}-\frac {1}{2 x}\right )}{x^{2}-1} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (\frac {x}{2}-\frac {1}{2 x}\right ) c_{1} -\frac {2 c_{2} \left (\frac {x}{2}-\frac {1}{2 x}\right )}{x^{2}-1} \\ \end{align*}
Verification of solutions
\[ y = \left (\frac {x}{2}-\frac {1}{2 x}\right ) c_{1} -\frac {2 c_{2} \left (\frac {x}{2}-\frac {1}{2 x}\right )}{x^{2}-1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+y^{\prime } x +y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {y^{\prime }}{x}-\frac {y}{x^{2}} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {y^{\prime }}{x}+\frac {y}{x^{2}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+y^{\prime } x +y=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & t =\ln \left (x \right ) \\ \square & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {1st}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (\frac {d}{d t}y \left (t \right )\right ) t^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\frac {d}{d t}y \left (t \right )}{x} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {2nd}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{2}+t^{\prime \prime }\left (x \right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}}\right )+\frac {d}{d t}y \left (t \right )+y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )+y \left (t \right )=0 \\ \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 (t \right )=\cos \left (t \right ) \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (t \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (t \right )=c_{1} \cos \left (t \right )+c_{2} \sin \left (t \right ) \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=c_{1} \cos \left (\ln \left (x \right )\right )+c_{2} \sin \left (\ln \left (x \right )\right ) \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type <- LODE of Euler type successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 15
dsolve([x^2*diff(y(x),x$2)+x*diff(y(x),x)+y(x)=0,sinh(ln(x))],singsol=all)
\[ y \left (x \right ) = c_{1} \sin \left (\ln \left (x \right )\right )+c_{2} \cos \left (\ln \left (x \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.022 (sec). Leaf size: 18
DSolve[x^2*y''[x]+x*y'[x]+y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 \cos (\log (x))+c_2 \sin (\log (x)) \]