Internal problem ID [1134]
Internal file name [OUTPUT/1135_Sunday_June_05_2022_02_03_18_AM_50290424/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 5 linear second order equations. Section 5.6 Reduction or order. Page
253
Problem number: 28.
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_2"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (1+2 x \right ) x y^{\prime \prime }-2 \left (2 x^{2}-1\right ) y^{\prime }-4 \left (x +1\right ) y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= \frac {1}{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 \[ p \left (x \right ) = \frac {-4 x^{2}+2}{2 x^{2}+x} \] Therefore \begin{align*} y_{2}\left (x \right ) &= \frac {\int {\mathrm e}^{-\left (\int \frac {-4 x^{2}+2}{2 x^{2}+x}d x \right )} x^{2}d x}{x} \\ y_{2}\left (x \right ) &= \frac {1}{x} \int \frac {{\mathrm e}^{2 x -2 \ln \left (x \right )+\ln \left (1+2 x \right )}}{\frac {1}{x^{2}}} , dx \\ y_{2}\left (x \right ) &= \frac {\int \left (1+2 x \right ) {\mathrm e}^{2 x}d x}{x} \\ y_{2}\left (x \right ) &= {\mathrm e}^{2 x} \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= \frac {c_{1}}{x}+c_{2} {\mathrm e}^{2 x} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1}}{x}+c_{2} {\mathrm e}^{2 x} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1}}{x}+c_{2} {\mathrm e}^{2 x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (2 x^{2}+x \right ) y^{\prime \prime }+\left (-4 x^{2}+2\right ) y^{\prime }+\left (-4 x -4\right ) 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 {4 \left (x +1\right ) y}{\left (1+2 x \right ) x}+\frac {2 \left (2 x^{2}-1\right ) y^{\prime }}{\left (1+2 x \right ) x} \\ \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 {2 \left (2 x^{2}-1\right ) y^{\prime }}{\left (1+2 x \right ) x}-\frac {4 \left (x +1\right ) y}{\left (1+2 x \right ) x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {2 \left (2 x^{2}-1\right )}{\left (1+2 x \right ) x}, P_{3}\left (x \right )=-\frac {4 \left (x +1\right )}{\left (1+2 x \right ) x}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=2 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (1+2 x \right ) x y^{\prime \prime }+\left (-4 x^{2}+2\right ) y^{\prime }+\left (-4 x -4\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \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..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (1+r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (2+r \right )+2 a_{0} \left (1+r \right ) \left (-2+r \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +r +1\right ) \left (k +2+r \right )+2 a_{k} \left (k +r +1\right ) \left (k +r -2\right )-4 a_{k -1} \left (k +r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 0\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (2+r \right )+2 a_{0} \left (1+r \right ) \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +r +1\right ) \left (k +2+r \right )+2 a_{k} \left (k +r +1\right ) \left (k +r -2\right )-4 a_{k -1} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (k +3+r \right )+2 a_{k +1} \left (k +2+r \right ) \left (k +r -1\right )-4 a_{k} \left (k +r +1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {2 \left (k^{2} a_{k +1}+2 k r a_{k +1}+r^{2} a_{k +1}-2 k a_{k}+k a_{k +1}-2 r a_{k}+r a_{k +1}-2 a_{k}-2 a_{k +1}\right )}{\left (k +2+r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=-\frac {2 \left (k^{2} a_{k +1}-2 k a_{k}-k a_{k +1}-2 a_{k +1}\right )}{\left (k +1\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}, a_{k +2}=-\frac {2 \left (k^{2} a_{k +1}-2 k a_{k}-k a_{k +1}-2 a_{k +1}\right )}{\left (k +1\right ) \left (k +2\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {2 \left (k^{2} a_{k +1}-2 k a_{k}+k a_{k +1}-2 a_{k}-2 a_{k +1}\right )}{\left (k +2\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=-\frac {2 \left (k^{2} a_{k +1}-2 k a_{k}+k a_{k +1}-2 a_{k}-2 a_{k +1}\right )}{\left (k +2\right ) \left (k +3\right )}, 2 a_{1}-4 a_{0}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k}\right ), a_{k +2}=-\frac {2 \left (k^{2} a_{1+k}-2 k a_{k}-k a_{1+k}-2 a_{1+k}\right )}{\left (1+k \right ) \left (k +2\right )}, 0=0, b_{k +2}=-\frac {2 \left (k^{2} b_{1+k}-2 k b_{k}+k b_{1+k}-2 b_{k}-2 b_{1+k}\right )}{\left (k +2\right ) \left (k +3\right )}, 2 b_{1}-4 b_{0}=0\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) Reducible group (found another exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 17
dsolve([(2*x+1)*x*diff(y(x),x$2)-2*(2*x^2-1)*diff(y(x),x)-4*(x+1)*y(x)=0,1/x],singsol=all)
\[ y \left (x \right ) = \frac {c_{2} {\mathrm e}^{2 x} x +c_{1}}{x} \]
✓ Solution by Mathematica
Time used: 0.049 (sec). Leaf size: 28
DSolve[(2*x+1)*x*y''[x]-2*(2*x^2-1)*y'[x]-4*(x+1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_2 e^{2 x+1} x+c_1}{\sqrt {e} x} \]