Internal problem ID [4861]
Internal file name [OUTPUT/4354_Sunday_June_05_2022_01_05_41_PM_95419094/index.tex]
Book: Mathematical Methods in the Physical Sciences. third edition. Mary L. Boas. John Wiley.
2006
Section: Chapter 8, Ordinary differential equations. Section 7. Other second-Order equations.
page 435
Problem number: 28.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "reduction_of_order"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {3 x y^{\prime \prime }-2 \left (-1+3 x \right ) y^{\prime }+\left (3 x -2\right ) y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= {\mathrm e}^{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 \[ p \left (x \right ) = \frac {-6 x +2}{3 x} \] Therefore \begin{align*} y_{2}\left (x \right ) &= {\mathrm e}^{x} \left (\int {\mathrm e}^{-\left (\int \frac {-6 x +2}{3 x}d x \right )} {\mathrm e}^{-2 x}d x \right ) \\ y_{2}\left (x \right ) &= {\mathrm e}^{x} \int \frac {{\mathrm e}^{2 x -\frac {2 \ln \left (x \right )}{3}}}{{\mathrm e}^{2 x}} , dx \\ y_{2}\left (x \right ) &= {\mathrm e}^{x} \left (\int \frac {1}{x^{\frac {2}{3}}}d x \right ) \\ y_{2}\left (x \right ) &= 3 \,{\mathrm e}^{x} x^{\frac {1}{3}} \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} {\mathrm e}^{x}+3 c_{2} {\mathrm e}^{x} x^{\frac {1}{3}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{x}+3 c_{2} {\mathrm e}^{x} x^{\frac {1}{3}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{x}+3 c_{2} {\mathrm e}^{x} x^{\frac {1}{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 x y^{\prime \prime }+\left (-6 x +2\right ) y^{\prime }+\left (3 x -2\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 {\left (3 x -2\right ) y}{3 x}+\frac {2 \left (-1+3 x \right ) y^{\prime }}{3 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 (-1+3 x \right ) y^{\prime }}{3 x}+\frac {\left (3 x -2\right ) y}{3 x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=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 (-1+3 x \right )}{3 x}, P_{3}\left (x \right )=\frac {3 x -2}{3 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}}}=\frac {2}{3} \\ {} & \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}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 3 x y^{\prime \prime }+\left (-6 x +2\right ) y^{\prime }+\left (3 x -2\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..1 \\ {} & {} & 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 \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-1+3 r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (2+3 r \right )-2 a_{0} \left (1+3 r \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (3 k +2+3 r \right )-2 a_{k} \left (3 k +3 r +1\right )+3 a_{k -1}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-1+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {1}{3}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (2+3 r \right )-2 a_{0} \left (1+3 r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 3 \left (k +\frac {2}{3}+r \right ) \left (k +1+r \right ) a_{k +1}-6 a_{k} k -6 a_{k} r -2 a_{k}+3 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 3 \left (k +\frac {5}{3}+r \right ) \left (k +2+r \right ) a_{k +2}-6 a_{k +1} \left (k +1\right )-6 r a_{k +1}-2 a_{k +1}+3 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {6 k a_{k +1}+6 r a_{k +1}-3 a_{k}+8 a_{k +1}}{\left (3 k +5+3 r \right ) \left (k +2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {6 k a_{k +1}-3 a_{k}+8 a_{k +1}}{\left (3 k +5\right ) \left (k +2\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 {6 k a_{k +1}-3 a_{k}+8 a_{k +1}}{\left (3 k +5\right ) \left (k +2\right )}, 2 a_{1}-2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & a_{k +2}=\frac {6 k a_{k +1}-3 a_{k}+10 a_{k +1}}{\left (3 k +6\right ) \left (k +\frac {7}{3}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{3}}, a_{k +2}=\frac {6 k a_{k +1}-3 a_{k}+10 a_{k +1}}{\left (3 k +6\right ) \left (k +\frac {7}{3}\right )}, 4 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}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{3}}\right ), a_{k +2}=\frac {6 k a_{k +1}-3 a_{k}+8 a_{k +1}}{\left (3 k +5\right ) \left (k +2\right )}, 2 a_{1}-2 a_{0}=0, b_{k +2}=\frac {6 k b_{k +1}-3 b_{k}+10 b_{k +1}}{\left (3 k +6\right ) \left (k +\frac {7}{3}\right )}, 4 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: 0.0 (sec). Leaf size: 14
dsolve([3*x*diff(y(x),x$2)-2*(3*x-1)*diff(y(x),x)+(3*x-2)*y(x)=0,exp(x)],singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{x} \left (c_{1} +x^{\frac {1}{3}} c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.026 (sec). Leaf size: 21
DSolve[3*x*y''[x]-2*(3*x-1)*y'[x]+(3*x-2)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^x \left (3 c_2 \sqrt [3]{x}+c_1\right ) \]