Internal problem ID [10878]
Internal file name [OUTPUT/10135_Sunday_December_24_2023_05_14_04_PM_3647982/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 2, Second-Order Differential Equations. section 2.1.2-2 Equation of form
\(y''+f(x)y'+g(x)y=0\)
Problem number: 55.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_change_of_variable_on_y_method_2"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }+\left (a \,x^{n}+x^{m} b \right ) y^{\prime }-\left (a \,x^{n -1}+b \,x^{m -1}\right ) y=0} \]
In normal form the ode \begin {align*} y^{\prime \prime }+\left (a \,x^{n}+x^{m} b \right ) y^{\prime }+\frac {\left (-a \,x^{n}-x^{m} b \right ) y}{x}&=0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=a \,x^{n}+x^{m} b\\ q \left (x \right )&=\frac {-a \,x^{n}-x^{m} b}{x} \end {align*}
Applying change of variables on the depndent variable \(y = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(y\). \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 (a \,x^{n}+x^{m} b \right )}{x}+\frac {-a \,x^{n}-x^{m} b}{x}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=1 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}+a \,x^{n}+x^{m} b \right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\left (\frac {2}{x}+a \,x^{n}+x^{m} b \right ) v^{\prime }\left (x \right )&=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 )+\left (\frac {2}{x}+a \,x^{n}+x^{m} b \right ) u \left (x \right ) = 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 {u \left (x^{n} a x +x^{m} b x +2\right )}{x} \end {align*}
Where \(f(x)=-\frac {x^{n} a x +x^{m} b x +2}{x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {x^{n} a x +x^{m} b x +2}{x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {x^{n} a x +x^{m} b x +2}{x} \,d x}\\ \ln \left (u \right )&=-2 \ln \left (x \right )-\frac {b x \,{\mathrm e}^{m \ln \left (x \right )}}{1+m}-\frac {a x \,{\mathrm e}^{n \ln \left (x \right )}}{n +1}+c_{1}\\ u&={\mathrm e}^{-2 \ln \left (x \right )-\frac {b x \,{\mathrm e}^{m \ln \left (x \right )}}{1+m}-\frac {a x \,{\mathrm e}^{n \ln \left (x \right )}}{n +1}+c_{1}}\\ &=c_{1} {\mathrm e}^{-2 \ln \left (x \right )-\frac {b x \,{\mathrm e}^{m \ln \left (x \right )}}{1+m}-\frac {a x \,{\mathrm e}^{n \ln \left (x \right )}}{n +1}} \end {align*}
Which simplifies to \[ u \left (x \right ) = \frac {c_{1} {\mathrm e}^{-\frac {b \,x^{1+m}}{1+m}} {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}}{x^{2}} \] 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}\\ &= \int \frac {c_{1} {\mathrm e}^{-\frac {b \,x^{1+m}}{1+m}} {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}}{x^{2}}d x +c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \left (\int \frac {c_{1} {\mathrm e}^{-\frac {b \,x^{1+m}}{1+m}} {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}}{x^{2}}d x +c_{2} \right ) x\\ &= \left (c_{1} \left (\int \frac {{\mathrm e}^{-\frac {\left (a \,x^{n} \left (1+m \right )+x^{m} b \left (n +1\right )\right ) x}{\left (n +1\right ) \left (1+m \right )}}}{x^{2}}d x \right )+c_{2} \right ) x\\ \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (\int \frac {c_{1} {\mathrm e}^{-\frac {b \,x^{1+m}}{1+m}} {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}}{x^{2}}d x +c_{2} \right ) x \\ \end{align*}
Verification of solutions
\[ y = \left (\int \frac {c_{1} {\mathrm e}^{-\frac {b \,x^{1+m}}{1+m}} {\mathrm e}^{-\frac {a \,x^{n +1}}{n +1}}}{x^{2}}d x +c_{2} \right ) x \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying an equivalence, under non-integer power transformations, to LODEs admitting Liouvillian solutions. -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) trying a symmetry of the form [xi=0, eta=F(x)] One independent solution has integrals. Trying a hypergeometric solution free of integrals... -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius No hypergeometric solution was found. <- linear_1 successful <- 2nd order, integrating factors of the form mu(x,y) successful`
✓ Solution by Maple
Time used: 0.328 (sec). Leaf size: 47
dsolve(diff(y(x),x$2)+(a*x^n+b*x^m)*diff(y(x),x)-(a*x^(n-1)+b*x^(m-1))*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = x \left (c_{1} +c_{2} \left (\int \frac {{\mathrm e}^{-\frac {\left (b \left (n +1\right ) x^{m}+a \left (1+m \right ) x^{n}\right ) x}{\left (1+m \right ) \left (n +1\right )}}}{x^{2}}d x \right )\right ) \]
✓ Solution by Mathematica
Time used: 1.216 (sec). Leaf size: 55
DSolve[y''[x]+(a*x^n+b*x^m)*y'[x]-(a*x^(n-1)+b*x^(m-1))*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x \left (c_2 \int _1^x\frac {\exp \left (K[1] \left (-\frac {b K[1]^m}{m+1}-\frac {a K[1]^n}{n+1}\right )\right )}{K[1]^2}dK[1]+c_1\right ) \]