Internal problem ID [11727]
Internal file name [OUTPUT/11737_Thursday_April_11_2024_08_49_15_PM_96516632/index.tex]
Book: Differential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi.
2004.
Section: Chapter 4, Section 4.1. Basic theory of linear differential equations. Exercises page
124
Problem number: 6.
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]]
\[ \boxed {\left (x^{3}-x^{2}\right ) y^{\prime \prime }-\left (x^{3}+2 x^{2}-2 x \right ) y^{\prime }+\left (2 x^{2}+2 x -2\right ) y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= 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 \[ p \left (x \right ) = \frac {-x^{3}-2 x^{2}+2 x}{x^{3}-x^{2}} \] Therefore \begin{align*} y_{2}\left (x \right ) &= x^{2} \left (\int \frac {{\mathrm e}^{-\left (\int \frac {-x^{3}-2 x^{2}+2 x}{x^{3}-x^{2}}d x \right )}}{x^{4}}d x \right ) \\ y_{2}\left (x \right ) &= x^{2} \int \frac {{\mathrm e}^{x +\ln \left (x -1\right )+2 \ln \left (x \right )}}{x^{4}} , dx \\ y_{2}\left (x \right ) &= x^{2} \left (\int \frac {\left (x -1\right ) {\mathrm e}^{x}}{x^{2}}d x \right ) \\ y_{2}\left (x \right ) &= {\mathrm e}^{x} x \\ \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} x^{2}+c_{2} {\mathrm e}^{x} x \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{2}+c_{2} {\mathrm e}^{x} x \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{2}+c_{2} {\mathrm e}^{x} x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x^{2} \left (x -1\right )+\left (-x^{3}-2 x^{2}+2 x \right ) y^{\prime }+\left (2 x^{2}+2 x -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 {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {2 \left (x^{2}+x -1\right ) y}{x^{2} \left (x -1\right )}+\frac {\left (x^{2}+2 x -2\right ) y^{\prime }}{x \left (x -1\right )} \\ \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 (x^{2}+2 x -2\right ) y^{\prime }}{x \left (x -1\right )}+\frac {2 \left (x^{2}+x -1\right ) y}{x^{2} \left (x -1\right )}=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 {x^{2}+2 x -2}{x \left (x -1\right )}, P_{3}\left (x \right )=\frac {2 \left (x^{2}+x -1\right )}{x^{2} \left (x -1\right )}\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}}}=2 \\ {} & \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 (\frac {d}{d x}y^{\prime }\right ) x^{2} \left (x -1\right )-x \left (x^{2}+2 x -2\right ) y^{\prime }+\left (2 x^{2}+2 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..2 \\ {} & {} & 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 =1..3 \\ {} & {} & 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 \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..3 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\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 \left (\frac {d}{d x}y^{\prime }\right )=\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} \left (-1+r \right ) \left (-2+r \right ) x^{r}+\left (-a_{1} r \left (-1+r \right )+a_{0} \left (-1+r \right ) \left (-2+r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (-a_{k} \left (k +r -1\right ) \left (k +r -2\right )+a_{k -1} \left (k +r -2\right ) \left (k -3+r \right )-a_{k -2} \left (k -4+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\left (-1+r \right ) \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{1, 2\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & -a_{1} r \left (-1+r \right )+a_{0} \left (-1+r \right ) \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=\frac {a_{0} \left (-2+r \right )}{r} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -a_{k} \left (k +r -1\right ) \left (k +r -2\right )+a_{k -1} \left (k +r -2\right ) \left (k -3+r \right )-a_{k -2} \left (k -4+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & -a_{k +2} \left (k +1+r \right ) \left (k +r \right )+a_{k +1} \left (k +r \right ) \left (k +r -1\right )-a_{k} \left (k +r -2\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}+2 k r a_{k +1}+r^{2} a_{k +1}-k a_{k}-k a_{k +1}-r a_{k}-r a_{k +1}+2 a_{k}}{\left (k +1+r \right ) \left (k +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}-k a_{k}+k a_{k +1}+a_{k}}{\left (k +2\right ) \left (k +1\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 {k^{2} a_{k +1}-k a_{k}+k a_{k +1}+a_{k}}{\left (k +2\right ) \left (k +1\right )}, a_{1}=-a_{0}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}-k a_{k}+3 k a_{k +1}+2 a_{k +1}}{\left (k +3\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +2}=\frac {k^{2} a_{k +1}-k a_{k}+3 k a_{k +1}+2 a_{k +1}}{\left (k +3\right ) \left (k +2\right )}, a_{1}=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 +2}\right ), a_{k +2}=\frac {k^{2} a_{k +1}-k a_{k}+k a_{k +1}+a_{k}}{\left (k +2\right ) \left (k +1\right )}, a_{1}=-a_{0}, b_{k +2}=\frac {k^{2} b_{k +1}-k b_{k}+3 k b_{k +1}+2 b_{k +1}}{\left (k +3\right ) \left (k +2\right )}, b_{1}=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([(x^3-x^2)*diff(y(x),x$2)-(x^3+2*x^2-2*x)*diff(y(x),x)+(2*x^2+2*x-2)*y(x)=0,x^2],singsol=all)
\[ y \left (x \right ) = x \left (c_{2} {\mathrm e}^{x}+c_{1} x \right ) \]
✓ Solution by Mathematica
Time used: 0.101 (sec). Leaf size: 36
DSolve[(x^3-x^2)*y''[x]-(x^3+2*x^2-2*x)*y'[x]+(2*x^2+2*x-2)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {\sqrt {1-x} x \left (c_2 x-c_1 e^x\right )}{\sqrt {x-1}} \]