Internal problem ID [10415]
Internal file name [OUTPUT/9363_Monday_June_06_2022_02_18_57_PM_81881186/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. subsection 1.2.3. Equations Containing
Exponential Functions
Problem number: 8.
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]
\[ \boxed {y^{\prime }-y^{2}=a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2}\), \(f_1(x)=0\) 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 &=0\\ f_2^2 f_0 &=a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\left (a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} {\mathrm e}^{-\frac {i {\mathrm e}^{4 \lambda x} a +4 \lambda ^{2} x \sqrt {a}+i {\mathrm e}^{2 \lambda x} b}{4 \lambda \sqrt {a}}} \operatorname {hypergeom}\left (\left [\frac {8 \lambda \,a^{\frac {3}{2}}+4 i c a -i b^{2}}{32 \lambda \,a^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{\frac {3}{2}}}\right )+c_{2} \operatorname {hypergeom}\left (\left [\frac {24 \lambda \,a^{\frac {3}{2}}+4 i c a -i b^{2}}{32 \lambda \,a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{\frac {3}{2}}}\right ) \left (2 a \,{\mathrm e}^{-\frac {-4 \lambda ^{2} x \sqrt {a}+i {\mathrm e}^{2 \lambda x} b +i {\mathrm e}^{4 \lambda x} a}{4 \lambda \sqrt {a}}}+{\mathrm e}^{-\frac {i {\mathrm e}^{4 \lambda x} a +4 \lambda ^{2} x \sqrt {a}+i {\mathrm e}^{2 \lambda x} b}{4 \lambda \sqrt {a}}} b \right ) \] 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
\[ \text {Expression too large to display} \] Warning, solution could not be verified
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}=a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2} \end {array} \]
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 Special trying Riccati sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (-a*exp(8*lambda*x)-b*exp(6*lambda*x)-c*exp(4*lambda*x)+lambda^2)*y(x) 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 -> 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 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 -> 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: indirect Equivalence to 0F1 under \`\`^ @ Moebius\`\` is resolved <- hypergeometric successful <- special function solution successful Change of variables used: [x = ln(t)/lambda] Linear ODE actually solved: (a*t^8+b*t^6+c*t^4-lambda^2)*u(t)+lambda^2*t*diff(u(t),t)+lambda^2*t^2*diff(diff(u(t),t),t) = 0 <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 1078
dsolve(diff(y(x),x)=y(x)^2+a*exp(8*lambda*x)+b*exp(6*lambda*x)+c*exp(4*lambda*x)-lambda^2,y(x), singsol=all)
\[ \text {Expression too large to display} \]
✓ Solution by Mathematica
Time used: 4.991 (sec). Leaf size: 1282
DSolve[y'[x]==y[x]^2+a*Exp[8*\[Lambda]*x]+b*Exp[6*\[Lambda]*x]+c*Exp[4*\[Lambda]*x]-\[Lambda]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {-e^{2 x \lambda } \operatorname {Hypergeometric1F1}\left (\frac {-i b^2+4 i a c+40 a^{3/2} \lambda }{32 a^{3/2} \lambda },\frac {3}{2},\frac {i \left (2 e^{2 x \lambda } a+b\right )^2}{8 a^{3/2} \lambda }\right ) b^3-2 a e^{4 x \lambda } \operatorname {Hypergeometric1F1}\left (\frac {-i b^2+4 i a c+40 a^{3/2} \lambda }{32 a^{3/2} \lambda },\frac {3}{2},\frac {i \left (2 e^{2 x \lambda } a+b\right )^2}{8 a^{3/2} \lambda }\right ) b^2+8 i a^{3/2} e^{2 x \lambda } \lambda \operatorname {Hypergeometric1F1}\left (\frac {-i b^2+4 i a c+8 a^{3/2} \lambda }{32 a^{3/2} \lambda },\frac {1}{2},\frac {i \left (2 e^{2 x \lambda } a+b\right )^2}{8 a^{3/2} \lambda }\right ) b+4 a c e^{2 x \lambda } \operatorname {Hypergeometric1F1}\left (\frac {-i b^2+4 i a c+40 a^{3/2} \lambda }{32 a^{3/2} \lambda },\frac {3}{2},\frac {i \left (2 e^{2 x \lambda } a+b\right )^2}{8 a^{3/2} \lambda }\right ) b-8 i a^{3/2} e^{2 x \lambda } \lambda \operatorname {Hypergeometric1F1}\left (\frac {-i b^2+4 i a c+40 a^{3/2} \lambda }{32 a^{3/2} \lambda },\frac {3}{2},\frac {i \left (2 e^{2 x \lambda } a+b\right )^2}{8 a^{3/2} \lambda }\right ) b+8 a^{3/2} \lambda \left (2 i e^{4 x \lambda } a+2 \lambda \sqrt {a}+i b e^{2 x \lambda }\right ) c_1 \operatorname {HermiteH}\left (\frac {i \left (b^2-4 a c+8 i a^{3/2} \lambda \right )}{16 a^{3/2} \lambda },\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \left (2 e^{2 x \lambda } a+b\right )}{a^{3/4} \sqrt {\lambda }}\right )+2 \sqrt [4]{-1} \sqrt {2} a^{3/4} e^{2 x \lambda } \sqrt {\lambda } \left (-i b^2+4 i a c+8 a^{3/2} \lambda \right ) c_1 \operatorname {HermiteH}\left (\frac {i \left (b^2-4 a c+24 i a^{3/2} \lambda \right )}{16 a^{3/2} \lambda },\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \left (2 e^{2 x \lambda } a+b\right )}{a^{3/4} \sqrt {\lambda }}\right )+16 a^2 \lambda ^2 \operatorname {Hypergeometric1F1}\left (\frac {-i b^2+4 i a c+8 a^{3/2} \lambda }{32 a^{3/2} \lambda },\frac {1}{2},\frac {i \left (2 e^{2 x \lambda } a+b\right )^2}{8 a^{3/2} \lambda }\right )+16 i a^{5/2} e^{4 x \lambda } \lambda \operatorname {Hypergeometric1F1}\left (\frac {-i b^2+4 i a c+8 a^{3/2} \lambda }{32 a^{3/2} \lambda },\frac {1}{2},\frac {i \left (2 e^{2 x \lambda } a+b\right )^2}{8 a^{3/2} \lambda }\right )+8 a^2 c e^{4 x \lambda } \operatorname {Hypergeometric1F1}\left (\frac {-i b^2+4 i a c+40 a^{3/2} \lambda }{32 a^{3/2} \lambda },\frac {3}{2},\frac {i \left (2 e^{2 x \lambda } a+b\right )^2}{8 a^{3/2} \lambda }\right )-16 i a^{5/2} e^{4 x \lambda } \lambda \operatorname {Hypergeometric1F1}\left (\frac {-i b^2+4 i a c+40 a^{3/2} \lambda }{32 a^{3/2} \lambda },\frac {3}{2},\frac {i \left (2 e^{2 x \lambda } a+b\right )^2}{8 a^{3/2} \lambda }\right )}{16 a^2 \lambda \left (c_1 \operatorname {HermiteH}\left (\frac {i \left (b^2-4 a c+8 i a^{3/2} \lambda \right )}{16 a^{3/2} \lambda },\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \left (2 e^{2 x \lambda } a+b\right )}{a^{3/4} \sqrt {\lambda }}\right )+\operatorname {Hypergeometric1F1}\left (\frac {-i b^2+4 i a c+8 a^{3/2} \lambda }{32 a^{3/2} \lambda },\frac {1}{2},\frac {i \left (2 e^{2 x \lambda } a+b\right )^2}{8 a^{3/2} \lambda }\right )\right )} \\ y(x)\to \frac {\left (\frac {1}{8}+\frac {i}{8}\right ) e^{2 \lambda x} \left (8 a^{3/2} \lambda +4 i a c-i b^2\right ) \operatorname {HermiteH}\left (\frac {i \left (b^2-4 a c+24 i a^{3/2} \lambda \right )}{16 a^{3/2} \lambda },\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \left (2 e^{2 x \lambda } a+b\right )}{a^{3/4} \sqrt {\lambda }}\right )}{a^{5/4} \sqrt {\lambda } \operatorname {HermiteH}\left (\frac {i \left (b^2-4 a c+8 i a^{3/2} \lambda \right )}{16 a^{3/2} \lambda },\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \left (2 e^{2 x \lambda } a+b\right )}{a^{3/4} \sqrt {\lambda }}\right )}+\frac {i b e^{2 \lambda x}}{2 \sqrt {a}}+i \sqrt {a} e^{4 \lambda x}+\lambda \\ \end{align*}