Internal problem ID [5526]
Internal file name [OUTPUT/4774_Sunday_June_05_2022_03_05_31_PM_62980122/index.tex]
Book: Notes on Diffy Qs. Differential Equations for Engineers. By by Jiri Lebl, 2013.
Section: Chapter 7. POWER SERIES METHODS. 7.3.2 The method of Frobenius. Exercises.
page 300
Problem number: 7.3.101 (b).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_bessel_ode", "second order series method. Irregular singular point"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {x^{3} y^{\prime \prime }+\left (1+x \right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ x^{3} y^{\prime \prime }+\left (1+x \right ) y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= 0\\ q(x) &= \frac {1+x}{x^{3}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([\infty ]\)
Irregular singular points : \([0]\)
Since \(x = 0\) is not an ordinary point, then we will now check if it is a regular singular point. Unable to solve since \(x = 0\) is not regular singular point. Terminating.
Verification of solutions N/A
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d}{d x}y^{\prime }\right )+\left (1+x \right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {\left (1+x \right ) y}{x^{3}} \\ \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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {\left (1+x \right ) y}{x^{3}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d}{d x}y^{\prime }\right )+\left (1+x \right ) 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} \\ {} & {} & \frac {d}{d x}y^{\prime }=\left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{2}+\left (\frac {d}{d x}t^{\prime }\left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {\frac {d}{d t}\frac {d}{d t}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^{3} \left (\frac {\frac {d}{d t}\frac {d}{d t}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}}\right )+\left (1+x \right ) y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )-\frac {d}{d t}y \left (t \right )\right )+\left (1+x \right ) y \left (t \right )=0 \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}\frac {d}{d t}y \left (t \right )=-\frac {\left (1+x \right ) y \left (t \right )}{x}+\frac {d}{d t}y \left (t \right ) \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (t \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d t}\frac {d}{d t}y \left (t \right )+\frac {\left (1+x \right ) y \left (t \right )}{x}-\frac {d}{d t}y \left (t \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+\frac {1+x}{x}-r =0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \frac {r^{2} x -r x +x +1}{x}=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (\frac {x +\sqrt {-3 x^{2}-4 x}}{2 x}, -\frac {-x +\sqrt {-3 x^{2}-4 x}}{2 x}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{\frac {\left (x +\sqrt {-3 x^{2}-4 x}\right ) t}{2 x}} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-\frac {\left (-x +\sqrt {-3 x^{2}-4 x}\right ) t}{2 x}} \\ \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} {\mathrm e}^{\frac {\left (x +\sqrt {-3 x^{2}-4 x}\right ) t}{2 x}}+c_{2} {\mathrm e}^{-\frac {\left (-x +\sqrt {-3 x^{2}-4 x}\right ) t}{2 x}} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=c_{1} {\mathrm e}^{\frac {\left (x +\sqrt {-3 x^{2}-4 x}\right ) \ln \left (x \right )}{2 x}}+c_{2} {\mathrm e}^{-\frac {\left (-x +\sqrt {-3 x^{2}-4 x}\right ) \ln \left (x \right )}{2 x}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=c_{1} x^{\frac {x +\sqrt {-3 x^{2}-4 x}}{2 x}}+c_{2} x^{-\frac {-x +\sqrt {-3 x^{2}-4 x}}{2 x}} \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 trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel <- Bessel successful <- special function solution successful`
✗ Solution by Maple
Order:=6; dsolve(x^3*diff(y(x),x$2)+(1+x)*y(x)=0,y(x),type='series',x=0);
\[ \text {No solution found} \]
✓ Solution by Mathematica
Time used: 0.036 (sec). Leaf size: 222
AsymptoticDSolveValue[x^3*y''[x]+(1+x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 e^{-\frac {2 i}{\sqrt {x}}} x^{3/4} \left (\frac {520667425699057 i x^{9/2}}{131941395333120}-\frac {21896102683 i x^{7/2}}{21474836480}+\frac {19100991 i x^{5/2}}{41943040}-\frac {3367 i x^{3/2}}{8192}-\frac {194208949785748261 x^5}{21110623253299200}+\frac {5189376335871 x^4}{2748779069440}-\frac {846810601 x^3}{1342177280}+\frac {205387 x^2}{524288}-\frac {273 x}{512}+\frac {13 i \sqrt {x}}{16}+1\right )+c_2 e^{\frac {2 i}{\sqrt {x}}} x^{3/4} \left (-\frac {520667425699057 i x^{9/2}}{131941395333120}+\frac {21896102683 i x^{7/2}}{21474836480}-\frac {19100991 i x^{5/2}}{41943040}+\frac {3367 i x^{3/2}}{8192}-\frac {194208949785748261 x^5}{21110623253299200}+\frac {5189376335871 x^4}{2748779069440}-\frac {846810601 x^3}{1342177280}+\frac {205387 x^2}{524288}-\frac {273 x}{512}-\frac {13 i \sqrt {x}}{16}+1\right ) \]