Internal problem ID [9470]
Internal file name [OUTPUT/8410_Monday_June_06_2022_02_58_56_AM_86973659/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1141.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {a x y^{\prime \prime }+\left (b x +3 a \right ) y^{\prime }+3 b y=0} \]
In normal form the ode \begin {align*} a x y^{\prime \prime }+\left (b x +3 a \right ) y^{\prime }+3 b y = 0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=r \left (x \right ) \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=\frac {b x +3 a}{a x}\\ q \left (x \right )&=\frac {3 b}{a x}\\ r \left (x \right )&=0 \end {align*}
The Lagrange adjoint ode is given by \begin {align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (\frac {\left (b x +3 a \right ) \xi \left (x \right )}{a x}\right )' + \left (\frac {3 b \xi \left (x \right )}{a x}\right ) &= 0\\ \xi ^{\prime \prime }\left (x \right )-\frac {\left (b x +3 a \right ) \xi ^{\prime }\left (x \right )}{a x}+\left (\frac {2 b}{a x}+\frac {b x +3 a}{a \,x^{2}}\right ) \xi \left (x \right )&= 0 \end {align*}
Which is solved for \(\xi (x)\). In normal form the ode \begin {align*} \xi ^{\prime \prime }\left (x \right ) a \,x^{2}+\left (-b \,x^{2}-3 a x \right ) \xi ^{\prime }\left (x \right )+\left (3 b x +3 a \right ) \xi \left (x \right )&=0 \tag {1} \end {align*}
Becomes \begin {align*} \xi ^{\prime \prime }\left (x \right )+p \left (x \right ) \xi ^{\prime }\left (x \right )+q \left (x \right ) \xi \left (x \right )&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=\frac {-b x -3 a}{a x}\\ q \left (x \right )&=\frac {3 b x +3 a}{a \,x^{2}} \end {align*}
Applying change of variables on the depndent variable \(\xi \left (x \right ) = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(\xi \left (x \right )\). \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2 n}{x}+p \right ) v^{\prime }\left (x \right )+\left (\frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q \right ) v \left (x \right )&=0 \tag {3} \end {align*}
Let the coefficient of \(v \left (x \right )\) above be zero. Hence \begin {align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q&=0 \tag {4} \end {align*}
Substituting the earlier values found for \(p \left (x \right )\) and \(q \left (x \right )\) into (4) gives \begin {align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n \left (-b x -3 a \right )}{a \,x^{2}}+\frac {3 b x +3 a}{a \,x^{2}}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=3 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {6}{x}+\frac {-b x -3 a}{a x}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\frac {\left (-b x +3 a \right ) v^{\prime }\left (x \right )}{a x}&=0 \tag {7} \\ \end {align*}
Using the substitution \begin {align*} u \left (x \right ) = v^{\prime }\left (x \right ) \end {align*}
Then (7) becomes \begin {align*} u^{\prime }\left (x \right )+\frac {\left (-b x +3 a \right ) u \left (x \right )}{a x} = 0 \tag {8} \\ \end {align*}
The above is now solved for \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {\left (-b x +3 a \right ) u}{a x} \end {align*}
Where \(f(x)=-\frac {-b x +3 a}{a x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {-b x +3 a}{a x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {-b x +3 a}{a x} \,d x}\\ \ln \left (u \right )&=-3 \ln \left (x \right )+\frac {x b}{a}+c_{1}\\ u&={\mathrm e}^{-3 \ln \left (x \right )+\frac {x b}{a}+c_{1}}\\ &=c_{1} {\mathrm e}^{-3 \ln \left (x \right )+\frac {x b}{a}} \end {align*}
Which simplifies to \[ u \left (x \right ) = \frac {c_{1} {\mathrm e}^{\frac {x b}{a}}}{x^{3}} \] Now that \(u \left (x \right )\) is known, then \begin {align*} v^{\prime }\left (x \right )&= u \left (x \right )\\ v \left (x \right )&= \int u \left (x \right )d x +c_{2}\\ &= \frac {c_{1} b^{2} \left (-\frac {a^{2} {\mathrm e}^{\frac {x b}{a}}}{2 x^{2} b^{2}}-\frac {{\mathrm e}^{\frac {x b}{a}} a}{2 x b}-\frac {\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right )}{2}\right )}{a^{2}}+c_{2} \end {align*}
Hence \begin {align*} \xi \left (x \right )&= v \left (x \right ) x^{n}\\ &= \left (\frac {c_{1} b^{2} \left (-\frac {a^{2} {\mathrm e}^{\frac {x b}{a}}}{2 x^{2} b^{2}}-\frac {{\mathrm e}^{\frac {x b}{a}} a}{2 x b}-\frac {\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right )}{2}\right )}{a^{2}}+c_{2} \right ) x^{3}\\ &= -\frac {\left (\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a b x -2 c_{2} a^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a^{2}\right ) x}{2 a^{2}}\\ \end {align*}
The original ode (2) now reduces to first order ode \begin {align*} \xi \left (x \right ) y^{\prime }-y \xi ^{\prime }\left (x \right )+\xi \left (x \right ) p \left (x \right ) y&=\int \xi \left (x \right ) r \left (x \right )d x\\ y^{\prime }+y \left (p \left (x \right )-\frac {\xi ^{\prime }\left (x \right )}{\xi \left (x \right )}\right )&=\frac {\int \xi \left (x \right ) r \left (x \right )d x}{\xi \left (x \right )}\\ y^{\prime }+y \left (\frac {b x +3 a}{a x}-\frac {c_{1} {\mathrm e}^{\frac {x b}{a}}+3 \left (\frac {c_{1} b^{2} \left (-\frac {a^{2} {\mathrm e}^{\frac {x b}{a}}}{2 x^{2} b^{2}}-\frac {{\mathrm e}^{\frac {x b}{a}} a}{2 x b}-\frac {\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right )}{2}\right )}{a^{2}}+c_{2} \right ) x^{2}}{\left (\frac {c_{1} b^{2} \left (-\frac {a^{2} {\mathrm e}^{\frac {x b}{a}}}{2 x^{2} b^{2}}-\frac {{\mathrm e}^{\frac {x b}{a}} a}{2 x b}-\frac {\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right )}{2}\right )}{a^{2}}+c_{2} \right ) x^{3}}\right )&=0 \end {align*}
Which is now a first order ode. This is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= -\frac {y \left (\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{3} x^{3}+{\mathrm e}^{\frac {x b}{a}} c_{1} a \,b^{2} x^{2}-2 c_{2} a^{2} b \,x^{3}+a^{2} x b c_{1} {\mathrm e}^{\frac {x b}{a}}+2 a^{3} c_{1} {\mathrm e}^{\frac {x b}{a}}\right )}{a x \left (\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a b x -2 c_{2} a^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a^{2}\right )} \end {align*}
Where \(f(x)=-\frac {\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{3} x^{3}+{\mathrm e}^{\frac {x b}{a}} c_{1} a \,b^{2} x^{2}-2 c_{2} a^{2} b \,x^{3}+a^{2} x b c_{1} {\mathrm e}^{\frac {x b}{a}}+2 a^{3} c_{1} {\mathrm e}^{\frac {x b}{a}}}{a x \left (\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a b x -2 c_{2} a^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a^{2}\right )}\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= -\frac {\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{3} x^{3}+{\mathrm e}^{\frac {x b}{a}} c_{1} a \,b^{2} x^{2}-2 c_{2} a^{2} b \,x^{3}+a^{2} x b c_{1} {\mathrm e}^{\frac {x b}{a}}+2 a^{3} c_{1} {\mathrm e}^{\frac {x b}{a}}}{a x \left (\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a b x -2 c_{2} a^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a^{2}\right )} \,d x\\ \int { \frac {1}{y} \,dy} &= \int {-\frac {\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{3} x^{3}+{\mathrm e}^{\frac {x b}{a}} c_{1} a \,b^{2} x^{2}-2 c_{2} a^{2} b \,x^{3}+a^{2} x b c_{1} {\mathrm e}^{\frac {x b}{a}}+2 a^{3} c_{1} {\mathrm e}^{\frac {x b}{a}}}{a x \left (\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a b x -2 c_{2} a^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a^{2}\right )} \,d x}\\ \ln \left (y \right )&=-2 \ln \left (x \right )+\ln \left (\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a b x -2 c_{2} a^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a^{2}\right )-\frac {x b}{a}+c_{3}\\ y&={\mathrm e}^{-2 \ln \left (x \right )+\ln \left (\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a b x -2 c_{2} a^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a^{2}\right )-\frac {x b}{a}+c_{3}}\\ &=c_{3} {\mathrm e}^{-2 \ln \left (x \right )+\ln \left (\operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a b x -2 c_{2} a^{2} x^{2}+{\mathrm e}^{\frac {x b}{a}} c_{1} a^{2}\right )-\frac {x b}{a}} \end {align*}
Which simplifies to \[ y = c_{3} \left ({\mathrm e}^{-\frac {x b}{a}} \operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{2}+\frac {c_{1} a b}{x}-2 \,{\mathrm e}^{-\frac {x b}{a}} c_{2} a^{2}+\frac {c_{1} a^{2}}{x^{2}}\right ) \] Hence, the solution found using Lagrange adjoint equation method is \[ y = c_{3} \left ({\mathrm e}^{-\frac {x b}{a}} \operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{2}+\frac {c_{1} a b}{x}-2 \,{\mathrm e}^{-\frac {x b}{a}} c_{2} a^{2}+\frac {c_{1} a^{2}}{x^{2}}\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} \left ({\mathrm e}^{-\frac {x b}{a}} \operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{2}+\frac {c_{1} a b}{x}-2 \,{\mathrm e}^{-\frac {x b}{a}} c_{2} a^{2}+\frac {c_{1} a^{2}}{x^{2}}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{3} \left ({\mathrm e}^{-\frac {x b}{a}} \operatorname {expIntegral}_{1}\left (-\frac {x b}{a}\right ) c_{1} b^{2}+\frac {c_{1} a b}{x}-2 \,{\mathrm e}^{-\frac {x b}{a}} c_{2} a^{2}+\frac {c_{1} a^{2}}{x^{2}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & a x \left (\frac {d}{d x}y^{\prime }\right )+\left (b x +3 a \right ) y^{\prime }+3 b 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 b y}{a x}-\frac {\left (b x +3 a \right ) y^{\prime }}{a 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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {\left (b x +3 a \right ) y^{\prime }}{a x}+\frac {3 b y}{a 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 {b x +3 a}{a x}, P_{3}\left (x \right )=\frac {3 b}{a 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}}}=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} \\ {} & {} & a x \left (\frac {d}{d x}y^{\prime }\right )+\left (b x +3 a \right ) y^{\prime }+3 b 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^{\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 \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\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 a_{0} r \left (2+r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a a_{k +1} \left (k +1+r \right ) \left (k +r +3\right )+b a_{k} \left (k +r +3\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & a r \left (2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-2, 0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r +3\right ) \left (a a_{k +1} \left (k +1+r \right )+b a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {b a_{k}}{a \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 \\ {} & {} & a_{k +1}=-\frac {b a_{k}}{a \left (k -1\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-2\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=-\frac {b a_{k}}{a \left (k -1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {b a_{k}}{a \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=-\frac {b a_{k}}{a \left (k +1\right )}\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) Group is reducible, not completely reducible <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 55
dsolve(a*x*diff(diff(y(x),x),x)+(b*x+3*a)*diff(y(x),x)+3*b*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{-\frac {b x}{a}} \operatorname {expIntegral}_{1}\left (-\frac {b x}{a}\right ) c_{2} b^{2} x^{2}+c_{1} {\mathrm e}^{-\frac {b x}{a}} x^{2}+c_{2} a \left (b x +a \right )}{x^{2}} \]
✓ Solution by Mathematica
Time used: 0.29 (sec). Leaf size: 63
DSolve[3*b*y[x] + (3*a + b*x)*y'[x] + a*x*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} \left (\frac {b^2 c_2 e^{-\frac {b x}{a}} \operatorname {ExpIntegralEi}\left (\frac {b x}{a}\right )}{a^2}-\frac {c_2 (a+b x)}{a x^2}+2 c_1 e^{-\frac {b x}{a}}\right ) \]