Internal problem ID [14464]
Internal file name [OUTPUT/14145_Monday_March_25_2024_09_50_26_PM_87594114/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.1, page 141
Problem number: 32.
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_1", "second_order_change_of_variable_on_y_method_2", "linear_second_order_ode_solved_by_an_integrating_factor"
Maple gives the following as the ode type
[[_Emden, _Fowler]]
\[ \boxed {t^{2} y^{\prime \prime }+6 t y^{\prime }+6 y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= \frac {1}{t^{2}} \end {align*}
Given one basis solution \(y_{1}\left (t \right )\), then the second basis solution is given by \[ y_{2}\left (t \right ) = y_{1} \left (\int \frac {{\mathrm e}^{-\left (\int p d t \right )}}{y_{1}^{2}}d t \right ) \] Where \(p(x)\) is the coefficient of \(y^{\prime }\) when the ode is written in the normal form \[ y^{\prime \prime }+p \left (t \right ) y^{\prime }+q \left (t \right ) y = f \left (t \right ) \] Looking at the ode to solve shows that \[ p \left (t \right ) = \frac {6}{t} \] Therefore \begin{align*} y_{2}\left (t \right ) &= \frac {\int {\mathrm e}^{-\left (\int \frac {6}{t}d t \right )} t^{4}d t}{t^{2}} \\ y_{2}\left (t \right ) &= \frac {1}{t^{2}} \int \frac {\frac {1}{t^{6}}}{\frac {1}{t^{4}}} , dt \\ y_{2}\left (t \right ) &= \frac {\int \frac {1}{t^{2}}d t}{t^{2}} \\ y_{2}\left (t \right ) &= -\frac {1}{t^{3}} \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= \frac {c_{1}}{t^{2}}-\frac {c_{2}}{t^{3}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1}}{t^{2}}-\frac {c_{2}}{t^{3}} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1}}{t^{2}}-\frac {c_{2}}{t^{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t^{2} y^{\prime \prime }+6 t y^{\prime }+6 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 {6 y^{\prime }}{t}-\frac {6 y}{t^{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 {6 y^{\prime }}{t}+\frac {6 y}{t^{2}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & t^{2} y^{\prime \prime }+6 t y^{\prime }+6 y=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & s =\ln \left (t \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 {t}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (\frac {d}{d s}y \left (s \right )\right ) s^{\prime }\left (t \right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\frac {d}{d s}y \left (s \right )}{t} \\ {} & \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 {t}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\left (\frac {d^{2}}{d s^{2}}y \left (s \right )\right ) {s^{\prime }\left (t \right )}^{2}+s^{\prime \prime }\left (t \right ) \left (\frac {d}{d s}y \left (s \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{2}}-\frac {\frac {d}{d s}y \left (s \right )}{t^{2}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & t^{2} \left (\frac {\frac {d^{2}}{d s^{2}}y \left (s \right )}{t^{2}}-\frac {\frac {d}{d s}y \left (s \right )}{t^{2}}\right )+6 \frac {d}{d s}y \left (s \right )+6 y \left (s \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d s^{2}}y \left (s \right )+5 \frac {d}{d s}y \left (s \right )+6 y \left (s \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+5 r +6=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +3\right ) \left (r +2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3, -2\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (s \right )={\mathrm e}^{-3 s} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (s \right )={\mathrm e}^{-2 s} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (s \right )=c_{1} y_{1}\left (s \right )+c_{2} y_{2}\left (s \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (s \right )=c_{1} {\mathrm e}^{-3 s}+c_{2} {\mathrm e}^{-2 s} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} s =\ln \left (t \right ) \\ {} & {} & y=\frac {c_{1}}{t^{3}}+\frac {c_{2}}{t^{2}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\frac {c_{1}}{t^{3}}+\frac {c_{2}}{t^{2}} \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: 13
dsolve([t^2*diff(y(t),t$2)+6*t*diff(y(t),t)+6*y(t)=0,1/t^2],singsol=all)
\[ y \left (t \right ) = \frac {c_{2} t +c_{1}}{t^{3}} \]
✓ Solution by Mathematica
Time used: 0.01 (sec). Leaf size: 16
DSolve[t^2*y''[t]+6*t*y'[t]+6*y[t]==0,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to \frac {c_2 t+c_1}{t^3} \]