Internal problem ID [6011]
Internal file name [OUTPUT/5259_Sunday_June_05_2022_03_28_47_PM_54326568/index.tex]
Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY
1961
Section: Chapter 3. Linear equations with variable coefficients. Page 121
Problem number: 1(c).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "reduction_of_order", "second_order_change_of_variable_on_y_method_1", "linear_second_order_ode_solved_by_an_integrating_factor"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }-4 y^{\prime } x +\left (4 x^{2}-2\right ) y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= {\mathrm e}^{x^{2}} \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 \[ p \left (x \right ) = -4 x \] Therefore \begin{align*} y_{2}\left (x \right ) &= {\mathrm e}^{x^{2}} \left (\int {\mathrm e}^{-\left (\int -4 x d x \right )} {\mathrm e}^{-2 x^{2}}d x \right ) \\ y_{2}\left (x \right ) &= {\mathrm e}^{x^{2}} \int \frac {{\mathrm e}^{2 x^{2}}}{{\mathrm e}^{2 x^{2}}} , dx \\ y_{2}\left (x \right ) &= {\mathrm e}^{x^{2}} \left (\int 1d x \right ) \\ y_{2}\left (x \right ) &= x \,{\mathrm e}^{x^{2}} \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= {\mathrm e}^{x^{2}} c_{1} +c_{2} x \,{\mathrm e}^{x^{2}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{x^{2}} c_{1} +c_{2} x \,{\mathrm e}^{x^{2}} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{x^{2}} c_{1} +c_{2} x \,{\mathrm e}^{x^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }-4 y^{\prime } x +\left (4 x^{2}-2\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 {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =\max \left (0, -m \right )}{\sum }}a_{k} x^{k +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =\max \left (0, -m \right )+m}{\sum }}a_{k -m} x^{k} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} k \,x^{k} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d}{d x}y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k} k \left (k -1\right ) x^{k -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k +2} \left (k +2\right ) \left (k +1\right ) x^{k} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 2 a_{2}-2 a_{0}+\left (6 a_{3}-6 a_{1}\right ) x +\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )-2 a_{k} \left (2 k +1\right )+4 a_{k -2}\right ) x^{k}\right )=0 \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [2 a_{2}-2 a_{0}=0, 6 a_{3}-6 a_{1}=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{2}=a_{0}, a_{3}=a_{1}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+3 k +2\right ) a_{k +2}-4 a_{k} k -2 a_{k}+4 a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (\left (k +2\right )^{2}+3 k +8\right ) a_{k +4}-4 a_{k +2} \left (k +2\right )-2 a_{k +2}+4 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines the series solution to the ODE}\hspace {3pt} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +4}=\frac {2 \left (2 k a_{k +2}-2 a_{k}+5 a_{k +2}\right )}{k^{2}+7 k +12}, a_{2}=a_{0}, a_{3}=a_{1}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`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 A Liouvillian solution exists Reducible group (found an exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 14
dsolve([diff(y(x),x$2)-4*x*diff(y(x),x)+(4*x^2-2)*y(x)=0,exp(x^2)],singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{x^{2}} \left (c_{2} x +c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.021 (sec). Leaf size: 18
DSolve[y''[x]-4*x*y'[x]+(4*x^2-2)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{x^2} (c_2 x+c_1) \]