Internal problem ID [11029]
Internal file name [OUTPUT/10286_Sunday_December_31_2023_04_01_02_PM_82853361/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 2.1.2-6 Equation of form
\((a_3 x^3+a_2 x^2 x+a_1 x+a_0) y''+f(x)y'+g(x)y=0\)
Problem number: 206.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_change_of_variable_on_x_method_1", "second_order_change_of_variable_on_x_method_2"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]
\[ \boxed {2 \left (a \,x^{3}+b \,x^{2}+c x +d \right ) y^{\prime \prime }+\left (3 a \,x^{2}+2 b x +c \right ) y^{\prime }+y \lambda =0} \]
In normal form the ode \begin {align*} y^{\prime \prime } \left (2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d \right )+\left (3 a \,x^{2}+2 b x +c \right ) y^{\prime }+y \lambda &=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 {3 a \,x^{2}+2 b x +c}{2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d}\\ q \left (x \right )&=\frac {\lambda }{2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d} \end {align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) gives \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}
Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}
Let \(p_{1} = 0\). Eq (4) simplifies to \begin {align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end {align*}
This ode is solved resulting in \begin {align*} \tau &= \int {\mathrm e}^{-\left (\int p \left (x \right )d x \right )}d x\\ &= \int {\mathrm e}^{-\left (\int \frac {3 a \,x^{2}+2 b x +c}{2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d}d x \right )}d x\\ &= \int e^{-\frac {\ln \left (a \,x^{3}+b \,x^{2}+c x +d \right )}{2}} \,dx\\ &= \int \frac {1}{\sqrt {a \,x^{3}+b \,x^{2}+c x +d}}d x\\ &= \text {Expression too large to display}\tag {6} \end {align*}
Using (6) to evaluate \(q_{1}\) from (5) gives \begin {align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {\frac {\lambda }{2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d}}{\frac {1}{a \,x^{3}+b \,x^{2}+c x +d}}\\ &= \frac {\lambda }{2}\tag {7} \end {align*}
Substituting the above in (3) and noting that now \(p_{1} = 0\) results in \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {\lambda y \left (\tau \right )}{2}&=0 \end {align*}
The above ode is now solved for \(y \left (\tau \right )\).This is second order with constant coefficients homogeneous ODE. In standard form the ODE is \[ A y''(\tau ) + B y'(\tau ) + C y(\tau ) = 0 \] Where in the above \(A=1, B=0, C=\frac {\lambda }{2}\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives \[ \lambda ^{2} {\mathrm e}^{\lambda \tau }+\frac {\lambda \,{\mathrm e}^{\lambda \tau }}{2} = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives \[ \lambda ^{2}+\frac {\lambda }{2} = 0 \tag {2} \] Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula \[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \] Substituting \(A=1, B=0, C=\frac {\lambda }{2}\) into the above gives \begin {align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (\frac {\lambda }{2}\right )}\\ &= \pm \frac {\sqrt {-2 \lambda }}{2} \end {align*}
Hence \begin{align*} \lambda _1 &= + \frac {\sqrt {-2 \lambda }}{2} \\ \lambda _2 &= - \frac {\sqrt {-2 \lambda }}{2} \\ \end{align*} Which simplifies to \begin{align*} \lambda _1 &= \frac {\sqrt {2}\, \sqrt {-\lambda }}{2} \\ \lambda _2 &= -\frac {\sqrt {2}\, \sqrt {-\lambda }}{2} \\ \end{align*} Since roots are real and distinct, then the solution is \begin{align*} y \left (\tau \right ) &= c_{1} e^{\lambda _1 \tau } + c_{2} e^{\lambda _2 \tau } \\ y \left (\tau \right ) &= c_{1} e^{\left (\frac {\sqrt {2}\, \sqrt {-\lambda }}{2}\right )\tau } +c_{2} e^{\left (-\frac {\sqrt {2}\, \sqrt {-\lambda }}{2}\right )\tau } \\ \end{align*} Or \[ y \left (\tau \right ) =c_{1} {\mathrm e}^{\frac {\sqrt {2}\, \sqrt {-\lambda }\, \tau }{2}}+c_{2} {\mathrm e}^{-\frac {\sqrt {2}\, \sqrt {-\lambda }\, \tau }{2}} \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= \text {Expression too large to display} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \text {Expression too large to display} \\ \end{align*}
Verification of solutions
\[ \text {Expression too large to display} \] Warning, solution could not be verified
In normal form the ode \begin {align*} y^{\prime \prime } \left (2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d \right )+\left (3 a \,x^{2}+2 b x +c \right ) y^{\prime }+y \lambda &=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 {3 a \,x^{2}+2 b x +c}{2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d}\\ q \left (x \right )&=\frac {\lambda }{2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d} \end {align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) results \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}
Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}
Let \(q_1=c^2\) where \(c\) is some constant. Therefore from (5) \begin {align*} \tau ' &= \frac {1}{c}\sqrt {q}\\ &= \frac {\sqrt {\frac {\lambda }{2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d}}}{c}\tag {6} \\ \tau '' &= -\frac {\lambda \left (6 a \,x^{2}+4 b x +2 c \right )}{2 c \sqrt {\frac {\lambda }{2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d}}\, \left (2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d \right )^{2}} \end {align*}
Substituting the above into (4) results in \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &=\frac {-\frac {\lambda \left (6 a \,x^{2}+4 b x +2 c \right )}{2 c \sqrt {\frac {\lambda }{2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d}}\, \left (2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d \right )^{2}}+\frac {3 a \,x^{2}+2 b x +c}{2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d}\frac {\sqrt {\frac {\lambda }{2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d}}}{c}}{\left (\frac {\sqrt {\frac {\lambda }{2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d}}}{c}\right )^2} \\ &=0 \end {align*}
Therefore ode (3) now becomes \begin {align*} y \left (\tau \right )'' + p_1 y \left (\tau \right )' + q_1 y \left (\tau \right ) &= 0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+c^{2} y \left (\tau \right ) &= 0 \tag {7} \end {align*}
The above ode is now solved for \(y \left (\tau \right )\). Since the ode is now constant coefficients, it can be easily solved to give \begin {align*} y \left (\tau \right ) &= c_{1} \cos \left (c \tau \right )+c_{2} \sin \left (c \tau \right ) \end {align*}
Now from (6) \begin {align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int \sqrt {\frac {\lambda }{2 a \,x^{3}+2 b \,x^{2}+2 c x +2 d}}d x}{c}\\ &= \text {Expression too large to display} \end {align*}
Substituting the above into the solution obtained gives \[ y = \text {Expression too large to display} \]
The solution(s) found are the following \begin{align*} \tag{1} \text {Expression too large to display} \\ \end{align*}
Verification of solutions
\[ \text {Expression too large to display} \] Warning, solution could not be verified
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)] Solution is available but has integrals. Trying a simpler solution using Kovacics algorithm... -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Group is reducible or imprimitive Solution has integrals. Trying a special function solution free of integrals... -> 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 No special function solution was found. <- Kovacics algorithm successful Solution via Kovacic is not simpler. Returning default solution <- linear_1 successful`
✓ Solution by Maple
Time used: 0.453 (sec). Leaf size: 67
dsolve(2*(a*x^3+b*x^2+c*x+d)*diff(y(x),x$2)+(3*a*x^2+2*b*x+c)*diff(y(x),x)+lambda*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} {\mathrm e}^{\frac {i \sqrt {2}\, \sqrt {\lambda }\, \left (\int \frac {1}{\sqrt {a \,x^{3}+x^{2} b +c x +d}}d x \right )}{2}}+c_{2} {\mathrm e}^{-\frac {i \sqrt {2}\, \sqrt {\lambda }\, \left (\int \frac {1}{\sqrt {a \,x^{3}+x^{2} b +c x +d}}d x \right )}{2}} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[2*(a*x^3+b*x^2+c*x+d)*y''[x]+(3*a*x^2+2*b*x+c)*y'[x]+lambda*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
Timed out