Internal problem ID [10356]
Internal file name [OUTPUT/9304_Monday_June_06_2022_01_50_23_PM_9360412/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: 27.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_Riccati]
\[ \boxed {y^{\prime }-y^{2}-\left (\alpha x +\beta \right ) y=x^{2} a +b x +c} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= x^{2} a +\alpha x y +b x +\beta y +y^{2}+c \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = x^{2} a +\alpha x y +b x +\beta y +y^{2}+c \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=x^{2} a +b x +c\), \(f_1(x)=\alpha x +\beta \) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u} \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' &=0\\ f_1 f_2 &=\alpha x +\beta \\ f_2^2 f_0 &=x^{2} a +b x +c \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )-\left (\alpha x +\beta \right ) u^{\prime }\left (x \right )+\left (x^{2} a +b x +c \right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = 4 \,{\mathrm e}^{\frac {\left (-\alpha ^{2} x +\alpha x \sqrt {\alpha ^{2}-4 a}+4 x a -2 \alpha \beta +2 \beta \sqrt {\alpha ^{2}-4 a}+4 b \right ) x}{4 \sqrt {\alpha ^{2}-4 a}}} \left (c_{2} \left (-\frac {1}{4} \alpha ^{2} x +x a -\frac {1}{4} \alpha \beta +\frac {1}{2} b \right ) \operatorname {hypergeom}\left (\left [\frac {\left (-\alpha ^{3}-2 \alpha ^{2} c +\left (2 \beta b +4 a \right ) \alpha +\left (-2 \beta ^{2}+8 c \right ) a -2 b^{2}\right ) \sqrt {\alpha ^{2}-4 a}+48 \left (-\frac {\alpha ^{2}}{4}+a \right )^{2}}{4 \left (-\alpha ^{2}+4 a \right )^{2}}\right ], \left [\frac {3}{2}\right ], \frac {\left (-\alpha ^{2} x +4 x a -\alpha \beta +2 b \right )^{2}}{2 \left (\alpha ^{2}-4 a \right )^{\frac {3}{2}}}\right )+\frac {\operatorname {hypergeom}\left (\left [\frac {\left (-\alpha ^{3}-2 \alpha ^{2} c +\left (2 \beta b +4 a \right ) \alpha +\left (-2 \beta ^{2}+8 c \right ) a -2 b^{2}\right ) \sqrt {\alpha ^{2}-4 a}+16 \left (-\frac {\alpha ^{2}}{4}+a \right )^{2}}{4 \left (-\alpha ^{2}+4 a \right )^{2}}\right ], \left [\frac {1}{2}\right ], \frac {\left (-\alpha ^{2} x +4 x a -\alpha \beta +2 b \right )^{2}}{2 \left (\alpha ^{2}-4 a \right )^{\frac {3}{2}}}\right ) c_{1}}{4}\right ) \] 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
\[ \text {Expression too large to display} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-\left (\alpha x +\beta \right ) y=x^{2} a +b x +c \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+\left (\alpha x +\beta \right ) y+x^{2} a +b x +c \end {array} \]
Maple trace
`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) = (alpha*x+beta)*(diff(y(x), x))+(-a*x^2-b*x-c)*y(x), y(x)` *** Sub 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 <- No Liouvillian solutions exists -> 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 <- hyper3 successful: indirect Equivalence to 0F1 under \`\`^ @ Moebius\`\` is resolved <- hypergeometric successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 974
dsolve(diff(y(x),x)=y(x)^2+(alpha*x+beta)*y(x)+a*x^2+b*x+c,y(x), singsol=all)
\[ \text {Expression too large to display} \]
✓ Solution by Mathematica
Time used: 4.264 (sec). Leaf size: 1291
DSolve[y'[x]==y[x]^2+(\[Alpha]*x+\[Beta])*y[x]+a*x^2+b*x+c,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {2 \left (2 b+4 a x+\left (\sqrt {\alpha ^2-4 a}-\alpha \right ) (x \alpha +\beta )\right ) \operatorname {Hypergeometric1F1}\left (-\frac {2 b^2-2 \alpha \beta b+\alpha ^2 \left (2 c+\alpha -\sqrt {\alpha ^2-4 a}\right )+2 a \left (\beta ^2-4 c-2 \alpha +2 \sqrt {\alpha ^2-4 a}\right )}{4 \left (\alpha ^2-4 a\right )^{3/2}},\frac {1}{2},\frac {(2 b+4 a x-\alpha (x \alpha +\beta ))^2}{2 \left (\alpha ^2-4 a\right )^{3/2}}\right ) \left (\alpha ^2-4 a\right )^2+2 (2 b+4 a x-\alpha (x \alpha +\beta )) \left (2 b^2-2 \alpha \beta b+\alpha ^2 \left (2 c+\alpha -\sqrt {\alpha ^2-4 a}\right )+2 a \left (\beta ^2-4 c-2 \alpha +2 \sqrt {\alpha ^2-4 a}\right )\right ) \operatorname {Hypergeometric1F1}\left (-\frac {2 b^2-2 \alpha \beta b+\alpha ^2 \left (2 c+\alpha -5 \sqrt {\alpha ^2-4 a}\right )+2 a \left (\beta ^2-4 c-2 \alpha +10 \sqrt {\alpha ^2-4 a}\right )}{4 \left (\alpha ^2-4 a\right )^{3/2}},\frac {3}{2},\frac {(2 b+4 a x-\alpha (x \alpha +\beta ))^2}{2 \left (\alpha ^2-4 a\right )^{3/2}}\right ) \sqrt {\alpha ^2-4 a}+\left (4 a-\alpha ^2\right ) c_1 \left (2 \left (4 a-\alpha ^2\right ) \left (2 b+4 a x+\left (\sqrt {\alpha ^2-4 a}-\alpha \right ) (x \alpha +\beta )\right ) \operatorname {HermiteH}\left (-\frac {-2 b^2+2 \alpha \beta b+\alpha ^2 \left (-2 c-\alpha +\sqrt {\alpha ^2-4 a}\right )-2 a \left (\beta ^2-4 c-2 \alpha +2 \sqrt {\alpha ^2-4 a}\right )}{2 \left (\alpha ^2-4 a\right )^{3/2}},\frac {-2 b-4 a x+\alpha (x \alpha +\beta )}{\sqrt {2} \left (\alpha ^2-4 a\right )^{3/4}}\right )-2 \sqrt {2} \sqrt [4]{\alpha ^2-4 a} \left (2 b^2-2 \alpha \beta b+\alpha ^2 \left (2 c+\alpha -\sqrt {\alpha ^2-4 a}\right )+2 a \left (\beta ^2-4 c-2 \alpha +2 \sqrt {\alpha ^2-4 a}\right )\right ) \operatorname {HermiteH}\left (\frac {2 b^2-2 \alpha \beta b+\alpha ^2 \left (2 c+\alpha -3 \sqrt {\alpha ^2-4 a}\right )+2 a \left (\beta ^2-4 c-2 \alpha +6 \sqrt {\alpha ^2-4 a}\right )}{2 \left (\alpha ^2-4 a\right )^{3/2}},\frac {-2 b-4 a x+\alpha (x \alpha +\beta )}{\sqrt {2} \left (\alpha ^2-4 a\right )^{3/4}}\right )\right )}{4 \left (\alpha ^2-4 a\right )^{5/2} \left (c_1 \operatorname {HermiteH}\left (-\frac {-2 b^2+2 \alpha \beta b+\alpha ^2 \left (-2 c-\alpha +\sqrt {\alpha ^2-4 a}\right )-2 a \left (\beta ^2-4 c-2 \alpha +2 \sqrt {\alpha ^2-4 a}\right )}{2 \left (\alpha ^2-4 a\right )^{3/2}},\frac {-2 b-4 a x+\alpha (x \alpha +\beta )}{\sqrt {2} \left (\alpha ^2-4 a\right )^{3/4}}\right )+\operatorname {Hypergeometric1F1}\left (-\frac {2 b^2-2 \alpha \beta b+\alpha ^2 \left (2 c+\alpha -\sqrt {\alpha ^2-4 a}\right )+2 a \left (\beta ^2-4 c-2 \alpha +2 \sqrt {\alpha ^2-4 a}\right )}{4 \left (\alpha ^2-4 a\right )^{3/2}},\frac {1}{2},\frac {(2 b+4 a x-\alpha (x \alpha +\beta ))^2}{2 \left (\alpha ^2-4 a\right )^{3/2}}\right )\right )} \\ y(x)\to \frac {\left (4 a-\alpha ^2\right ) \left (\left (\sqrt {\alpha ^2-4 a}-\alpha \right ) (\beta +\alpha x)+4 a x+2 b\right )-\frac {\sqrt {2} \sqrt [4]{\alpha ^2-4 a} \left (2 a \left (2 \sqrt {\alpha ^2-4 a}-2 \alpha +\beta ^2-4 c\right )+\alpha ^2 \left (-\sqrt {\alpha ^2-4 a}+\alpha +2 c\right )+2 b^2-2 \alpha \beta b\right ) \operatorname {HermiteH}\left (\frac {2 b^2-2 \alpha \beta b+\alpha ^2 \left (2 c+\alpha -3 \sqrt {\alpha ^2-4 a}\right )+2 a \left (\beta ^2-4 c-2 \alpha +6 \sqrt {\alpha ^2-4 a}\right )}{2 \left (\alpha ^2-4 a\right )^{3/2}},\frac {-2 b-4 a x+\alpha (x \alpha +\beta )}{\sqrt {2} \left (\alpha ^2-4 a\right )^{3/4}}\right )}{\operatorname {HermiteH}\left (-\frac {-2 b^2+2 \alpha \beta b+\alpha ^2 \left (-2 c-\alpha +\sqrt {\alpha ^2-4 a}\right )-2 a \left (\beta ^2-4 c-2 \alpha +2 \sqrt {\alpha ^2-4 a}\right )}{2 \left (\alpha ^2-4 a\right )^{3/2}},\frac {-2 b-4 a x+\alpha (x \alpha +\beta )}{\sqrt {2} \left (\alpha ^2-4 a\right )^{3/4}}\right )}}{2 \left (\alpha ^2-4 a\right )^{3/2}} \\ \end{align*}