Internal problem ID [11723]
Internal file name [OUTPUT/11733_Thursday_April_11_2024_08_49_14_PM_11953513/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: 2.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "reduction_of_order", "second_order_change_of_variable_on_x_method_1", "second_order_change_of_variable_on_x_method_2", "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (x +1\right )^{2} y^{\prime \prime }-3 \left (x +1\right ) y^{\prime }+3 y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= x +1 \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 \[ p \left (x \right ) = \frac {-3 x -3}{x^{2}+2 x +1} \] Therefore \begin{align*} y_{2}\left (x \right ) &= \left (x +1\right ) \left (\int \frac {{\mathrm e}^{-\left (\int \frac {-3 x -3}{x^{2}+2 x +1}d x \right )}}{\left (x +1\right )^{2}}d x \right ) \\ y_{2}\left (x \right ) &= x +1 \int \frac {\left (x +1\right )^{3}}{\left (x +1\right )^{2}} , dx \\ y_{2}\left (x \right ) &= \left (x +1\right ) \left (\int \left (x +1\right )d x \right ) \\ y_{2}\left (x \right ) &= \left (x +1\right ) \left (\frac {1}{2} x^{2}+x \right ) \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= \left (x +1\right ) c_{1} +c_{2} \left (x +1\right ) \left (\frac {1}{2} x^{2}+x \right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (x +1\right ) c_{1} +c_{2} \left (x +1\right ) \left (\frac {1}{2} x^{2}+x \right ) \\ \end{align*}
Verification of solutions
\[ y = \left (x +1\right ) c_{1} +c_{2} \left (x +1\right ) \left (\frac {1}{2} x^{2}+x \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x +1\right )^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (-3 x -3\right ) y^{\prime }+3 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 {3 y^{\prime }}{x +1}-\frac {3 y}{\left (x +1\right )^{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} \\ {} & {} & \frac {d}{d x}y^{\prime }-\frac {3 y^{\prime }}{x +1}+\frac {3 y}{\left (x +1\right )^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-1\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {3}{x +1}, P_{3}\left (x \right )=\frac {3}{\left (x +1\right )^{2}}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=-3 \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=3 \\ {} & \circ & x =-1\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-1\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (x +1\right )^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (-3 x -3\right ) y^{\prime }+3 y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & u^{2} \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )-3 u \left (\frac {d}{d u}y \left (u \right )\right )+3 y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite DE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u \cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u \cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{2}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u^{2}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r} \\ & {} & \textrm {Rewrite DE with series expansions}\hspace {3pt} \\ {} & {} & \moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r -1\right ) \left (k +r -3\right ) u^{k +r}=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k -1\right ) \left (k -3\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k}=0 \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k}=0\right ] \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: 17
dsolve([(x+1)^2*diff(y(x),x$2)-3*(x+1)*diff(y(x),x)+3*y(x)=0,x+1],singsol=all)
\[ y \left (x \right ) = \left (1+x \right ) \left (c_{1} +c_{2} \left (1+x \right )^{2}\right ) \]
✓ Solution by Mathematica
Time used: 0.026 (sec). Leaf size: 20
DSolve[(x+1)^2*y''[x]-3*(x+1)*y'[x]+3*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 (x+1)^3+c_1 (x+1) \]