Internal problem ID [10390]
Internal file name [OUTPUT/9338_Monday_June_06_2022_02_01_30_PM_6267541/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 1, section 1.2. Riccati Equation. 1.2.2. Equations Containing Power
Functions
Problem number: 61.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_rational, _Riccati]
\[ \boxed {\left (a_{2} x^{2}+b_{2} x +c_{2} \right ) y^{\prime }-y^{2}-\left (a_{1} x +b_{1} \right ) y=-\lambda \left (\lambda +a_{1} -a_{2} \right ) x^{2}+\lambda \left (b_{2} -b_{1} \right ) x +\lambda c_{2}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {-\lambda \,x^{2} a_{1} +\lambda \,x^{2} a_{2} -\lambda ^{2} x^{2}+a_{1} x y -\lambda x b_{1} +\lambda x b_{2} +b_{1} y +\lambda c_{2} +y^{2}}{a_{2} x^{2}+b_{2} x +c_{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\frac {\lambda \,x^{2} a_{1}}{a_{2} x^{2}+b_{2} x +c_{2}}+\frac {\lambda \,x^{2} a_{2}}{a_{2} x^{2}+b_{2} x +c_{2}}-\frac {\lambda ^{2} x^{2}}{a_{2} x^{2}+b_{2} x +c_{2}}+\frac {a_{1} x y}{a_{2} x^{2}+b_{2} x +c_{2}}-\frac {\lambda x b_{1}}{a_{2} x^{2}+b_{2} x +c_{2}}+\frac {\lambda x b_{2}}{a_{2} x^{2}+b_{2} x +c_{2}}+\frac {\lambda c_{2}}{a_{2} x^{2}+b_{2} x +c_{2}}+\frac {b_{1} y}{a_{2} x^{2}+b_{2} x +c_{2}}+\frac {y^{2}}{a_{2} x^{2}+b_{2} x +c_{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {-\lambda \,x^{2} a_{1} +\lambda \,x^{2} a_{2} -\lambda ^{2} x^{2}-\lambda x b_{1} +\lambda x b_{2} +\lambda c_{2}}{a_{2} x^{2}+b_{2} x +c_{2}}\), \(f_1(x)=\frac {a_{1} x +b_{1}}{a_{2} x^{2}+b_{2} x +c_{2}}\) and \(f_2(x)=\frac {1}{a_{2} x^{2}+b_{2} x +c_{2}}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{a_{2} x^{2}+b_{2} x +c_{2}}} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=-\frac {2 a_{2} x +b_{2}}{\left (a_{2} x^{2}+b_{2} x +c_{2} \right )^{2}}\\ f_1 f_2 &=\frac {a_{1} x +b_{1}}{\left (a_{2} x^{2}+b_{2} x +c_{2} \right )^{2}}\\ f_2^2 f_0 &=\frac {-\lambda \,x^{2} a_{1} +\lambda \,x^{2} a_{2} -\lambda ^{2} x^{2}-\lambda x b_{1} +\lambda x b_{2} +\lambda c_{2}}{\left (a_{2} x^{2}+b_{2} x +c_{2} \right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {u^{\prime \prime }\left (x \right )}{a_{2} x^{2}+b_{2} x +c_{2}}-\left (-\frac {2 a_{2} x +b_{2}}{\left (a_{2} x^{2}+b_{2} x +c_{2} \right )^{2}}+\frac {a_{1} x +b_{1}}{\left (a_{2} x^{2}+b_{2} x +c_{2} \right )^{2}}\right ) u^{\prime }\left (x \right )+\frac {\left (-\lambda \,x^{2} a_{1} +\lambda \,x^{2} a_{2} -\lambda ^{2} x^{2}-\lambda x b_{1} +\lambda x b_{2} +\lambda c_{2} \right ) u \left (x \right )}{\left (a_{2} x^{2}+b_{2} x +c_{2} \right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ \text {Expression too large to display} \] The above shows that \[ \text {Expression too large to display} \] Using the above in (1) gives the solution \[ \text {Expression too large to display} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ \text {Expression too large to display} \]
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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a_{2} x^{2}+b_{2} x +c_{2} \right ) y^{\prime }-y^{2}-\left (a_{1} x +b_{1} \right ) y=-\lambda \left (\lambda +a_{1} -a_{2} \right ) x^{2}+\lambda \left (b_{2} -b_{1} \right ) x +\lambda c_{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2}+\left (a_{1} x +b_{1} \right ) y-\lambda \left (\lambda +a_{1} -a_{2} \right ) x^{2}+\lambda \left (b_{2} -b_{1} \right ) x +\lambda c_{2}}{a_{2} x^{2}+b_{2} x +c_{2}} \end {array} \]
Maple trace Kovacic algorithm successful
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying Riccati trying Riccati sub-methods: trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (a__1*x-2*a__2*x+b__1-b__2)*(diff(y(x), x))/(a__2*x^2+b__2*x+c__2)-lam 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 Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric successful <- special function solution successful -> Trying to convert hypergeometric functions to elementary form... <- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integrals <- Kovacics algorithm successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 5771
dsolve((a__2*x^2+b__2*x+c__2)*diff(y(x),x)=y(x)^2+(a__1*x+b__1)*y(x)-lambda*(lambda+a__1-a__2)*x^2+lambda*(b__2-b__1)*x+lambda*c__2,y(x), singsol=all)
\[ \text {Expression too large to display} \]
✓ Solution by Mathematica
Time used: 34.34 (sec). Leaf size: 284
DSolve[(a2*x^2+b2*x+c2)*y'[x]==y[x]^2+(a1*x+b1)*y[x]-\[Lambda]*(\[Lambda]+a1-a2)*x^2+\[Lambda]*(b2-b1)*x+\[Lambda]*c2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\lambda x \int _1^x\exp \left (\frac {(\text {a1}-2 \text {a2}+2 \lambda ) \log (\text {c2}+K[1] (\text {b2}+\text {a2} K[1]))-\frac {2 (\text {b2} (\text {a1}+2 \lambda )-2 \text {a2} \text {b1}) \arctan \left (\frac {\text {b2}+2 \text {a2} K[1]}{\sqrt {4 \text {a2} \text {c2}-\text {b2}^2}}\right )}{\sqrt {4 \text {a2} \text {c2}-\text {b2}^2}}}{2 \text {a2}}\right )dK[1]+(x (\text {a2} x+\text {b2})+\text {c2})^{\frac {\text {a1}+2 \lambda }{2 \text {a2}}} \left (-\exp \left (\frac {(2 \text {a2} \text {b1}-\text {b2} (\text {a1}+2 \lambda )) \arctan \left (\frac {2 \text {a2} x+\text {b2}}{\sqrt {4 \text {a2} \text {c2}-\text {b2}^2}}\right )}{\text {a2} \sqrt {4 \text {a2} \text {c2}-\text {b2}^2}}\right )\right )+c_1 \lambda x}{\int _1^x\exp \left (\frac {(\text {a1}-2 \text {a2}+2 \lambda ) \log (\text {c2}+K[1] (\text {b2}+\text {a2} K[1]))-\frac {2 (\text {b2} (\text {a1}+2 \lambda )-2 \text {a2} \text {b1}) \arctan \left (\frac {\text {b2}+2 \text {a2} K[1]}{\sqrt {4 \text {a2} \text {c2}-\text {b2}^2}}\right )}{\sqrt {4 \text {a2} \text {c2}-\text {b2}^2}}}{2 \text {a2}}\right )dK[1]+c_1} \\ y(x)\to \lambda x \\ \end{align*}