Internal problem ID [9718]
Internal file name [OUTPUT/8660_Monday_June_06_2022_04_41_56_AM_259911/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1391.
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]]
\[ \boxed {y^{\prime \prime }-\frac {\left (7 a \,x^{2}+5\right ) y^{\prime }}{x \left (a \,x^{2}+1\right )}+\frac {\left (15 a \,x^{2}+5\right ) y}{x^{2} \left (a \,x^{2}+1\right )}=0} \]
In normal form the ode \begin {align*} \left (a \,x^{4}+x^{2}\right ) y^{\prime \prime }+\left (-7 a \,x^{3}-5 x \right ) y^{\prime }+\left (15 a \,x^{2}+5\right ) y&=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 )&=\frac {-7 a \,x^{2}-5}{x \left (a \,x^{2}+1\right )}\\ q \left (x \right )&=\frac {15 a \,x^{2}+5}{x^{2} \left (a \,x^{2}+1\right )} \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 (-7 a \,x^{2}-5\right )}{x^{2} \left (a \,x^{2}+1\right )}+\frac {15 a \,x^{2}+5}{x^{2} \left (a \,x^{2}+1\right )}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=5 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {10}{x}+\frac {-7 a \,x^{2}-5}{x \left (a \,x^{2}+1\right )}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\frac {\left (3 a \,x^{2}+5\right ) v^{\prime }\left (x \right )}{x \left (a \,x^{2}+1\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 )+\frac {\left (3 a \,x^{2}+5\right ) u \left (x \right )}{x \left (a \,x^{2}+1\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 {\left (3 a \,x^{2}+5\right ) u}{x \left (a \,x^{2}+1\right )} \end {align*}
Where \(f(x)=-\frac {3 a \,x^{2}+5}{x \left (a \,x^{2}+1\right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {3 a \,x^{2}+5}{x \left (a \,x^{2}+1\right )} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {3 a \,x^{2}+5}{x \left (a \,x^{2}+1\right )} \,d x}\\ \ln \left (u \right )&=\ln \left (a \,x^{2}+1\right )-5 \ln \left (x \right )+c_{1}\\ u&={\mathrm e}^{\ln \left (a \,x^{2}+1\right )-5 \ln \left (x \right )+c_{1}}\\ &=c_{1} {\mathrm e}^{\ln \left (a \,x^{2}+1\right )-5 \ln \left (x \right )} \end {align*}
Which simplifies to \[ u \left (x \right ) = c_{1} \left (\frac {a}{x^{3}}+\frac {1}{x^{5}}\right ) \] 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}\\ &= c_{1} \left (-\frac {1}{4 x^{4}}-\frac {a}{2 x^{2}}\right )+c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \left (c_{1} \left (-\frac {1}{4 x^{4}}-\frac {a}{2 x^{2}}\right )+c_{2} \right ) x^{5}\\ &= -\frac {\left (-4 c_{2} x^{4}+2 a c_{1} x^{2}+c_{1} \right ) x}{4}\\ \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{1} \left (-\frac {1}{4 x^{4}}-\frac {a}{2 x^{2}}\right )+c_{2} \right ) x^{5} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{1} \left (-\frac {1}{4 x^{4}}-\frac {a}{2 x^{2}}\right )+c_{2} \right ) x^{5} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a \,x^{4}+x^{2}\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-7 a \,x^{3}-5 x \right ) y^{\prime }+\left (15 a \,x^{2}+5\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 {5 \left (3 a \,x^{2}+1\right ) y}{x^{2} \left (a \,x^{2}+1\right )}+\frac {\left (7 a \,x^{2}+5\right ) y^{\prime }}{x \left (a \,x^{2}+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 (7 a \,x^{2}+5\right ) y^{\prime }}{x \left (a \,x^{2}+1\right )}+\frac {5 \left (3 a \,x^{2}+1\right ) y}{x^{2} \left (a \,x^{2}+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 {7 a \,x^{2}+5}{x \left (a \,x^{2}+1\right )}, P_{3}\left (x \right )=\frac {5 \left (3 a \,x^{2}+1\right )}{x^{2} \left (a \,x^{2}+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}}}=-5 \\ {} & \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}}}=5 \\ {} & \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 (a \,x^{2}+1\right )-x \left (7 a \,x^{2}+5\right ) y^{\prime }+\left (15 a \,x^{2}+5\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..4 \\ {} & {} & 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 (-5+r \right ) x^{r}+a_{1} r \left (-4+r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +r -1\right ) \left (k +r -5\right )+a a_{k -2} \left (k +r -5\right ) \left (k -7+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 (-5+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{1, 5\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} r \left (-4+r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r -5\right ) \left (a_{k} \left (k +r -1\right )+a a_{k -2} \left (k -7+r \right )\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (k +r -3\right ) \left (a_{k +2} \left (k +1+r \right )+a a_{k} \left (k +r -5\right )\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a a_{k} \left (k +r -5\right )}{k +1+r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =4 \\ {} & {} & a_{k +2}=-\frac {a a_{k} \left (k -4\right )}{k +2} \\ \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 {a a_{k} \left (k -4\right )}{k +2}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =5 \\ {} & {} & a_{k +2}=-\frac {a a_{k} k}{k +6} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =5 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +5}, a_{k +2}=-\frac {a a_{k} k}{k +6}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +5}\right ), b_{k +2}=-\frac {a b_{k} \left (k -4\right )}{k +2}, b_{1}=0, c_{k +2}=-\frac {a c_{k} k}{k +6}, c_{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) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 20
dsolve(diff(diff(y(x),x),x) = 1/x*(7*a*x^2+5)/(a*x^2+1)*diff(y(x),x)-(15*a*x^2+5)/x^2/(a*x^2+1)*y(x),y(x), singsol=all)
\[ y \left (x \right ) = c_{1} x^{5}+2 a c_{2} x^{3}+x c_{2} \]
✓ Solution by Mathematica
Time used: 0.064 (sec). Leaf size: 27
DSolve[y''[x] == -(((5 + 15*a*x^2)*y[x])/(x^2*(1 + a*x^2))) + ((5 + 7*a*x^2)*y'[x])/(x*(1 + a*x^2)),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 x^5-\frac {1}{4} c_2 x \left (2 a x^2+1\right ) \]