Internal problem ID [10453]
Internal file name [OUTPUT/9401_Monday_June_06_2022_02_22_19_PM_6503264/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 1.2.4-1. Equations with hyperbolic
sine and cosine
Problem number: 6.
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]
Unable to solve or complete the solution.
\[ \boxed {\left (a \sinh \left (\lambda x \right )+b \right ) y^{\prime }-y^{2}-c \sinh \left (\mu x \right ) y=-d^{2}+c d \sinh \left (\mu x \right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y^{2}+c \sinh \left (\mu x \right ) y -d^{2}+c d \sinh \left (\mu x \right )}{a \sinh \left (\lambda x \right )+b} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {c d \sinh \left (\mu x \right )}{a \sinh \left (\lambda x \right )+b}+\frac {c \sinh \left (\mu x \right ) y}{a \sinh \left (\lambda x \right )+b}-\frac {d^{2}}{a \sinh \left (\lambda x \right )+b}+\frac {y^{2}}{a \sinh \left (\lambda x \right )+b} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {-d^{2}+c d \sinh \left (\mu x \right )}{a \sinh \left (\lambda x \right )+b}\), \(f_1(x)=\frac {c \sinh \left (\mu x \right )}{a \sinh \left (\lambda x \right )+b}\) and \(f_2(x)=\frac {1}{a \sinh \left (\lambda x \right )+b}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{a \sinh \left (\lambda x \right )+b}} \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 {a \lambda \cosh \left (\lambda x \right )}{\left (a \sinh \left (\lambda x \right )+b \right )^{2}}\\ f_1 f_2 &=\frac {c \sinh \left (\mu x \right )}{\left (a \sinh \left (\lambda x \right )+b \right )^{2}}\\ f_2^2 f_0 &=\frac {-d^{2}+c d \sinh \left (\mu x \right )}{\left (a \sinh \left (\lambda x \right )+b \right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {u^{\prime \prime }\left (x \right )}{a \sinh \left (\lambda x \right )+b}-\left (-\frac {a \lambda \cosh \left (\lambda x \right )}{\left (a \sinh \left (\lambda x \right )+b \right )^{2}}+\frac {c \sinh \left (\mu x \right )}{\left (a \sinh \left (\lambda x \right )+b \right )^{2}}\right ) u^{\prime }\left (x \right )+\frac {\left (-d^{2}+c d \sinh \left (\mu x \right )\right ) u \left (x \right )}{\left (a \sinh \left (\lambda x \right )+b \right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives Unable to solve. Terminating.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a \sinh \left (\lambda x \right )+b \right ) y^{\prime }-y^{2}-c \sinh \left (\mu x \right ) y=-d^{2}+c d \sinh \left (\mu x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2}+c \sinh \left (\mu x \right ) y-d^{2}+c d \sinh \left (\mu x \right )}{a \sinh \left (\lambda x \right )+b} \end {array} \]
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: <- Riccati particular case Kamke (b) successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 253
dsolve((a*sinh(lambda*x)+b)*diff(y(x),x)=y(x)^2+c*sinh(mu*x)*y(x)-d^2+c*d*sinh(mu*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-d \left (\int \frac {{\mathrm e}^{\frac {c \left (\int \frac {\sinh \left (x \mu \right )}{\sinh \left (x \lambda \right ) a +b}d x \right ) \lambda \sqrt {a^{2}+b^{2}}+4 d \,\operatorname {arctanh}\left (\frac {-\tanh \left (\frac {x \lambda }{2}\right ) b +a}{\sqrt {a^{2}+b^{2}}}\right )}{\lambda \sqrt {a^{2}+b^{2}}}}}{\sinh \left (x \lambda \right ) a +b}d x \right )+d c_{1} -{\mathrm e}^{\frac {c \left (\int \frac {\sinh \left (x \mu \right )}{\sinh \left (x \lambda \right ) a +b}d x \right ) \lambda \sqrt {a^{2}+b^{2}}+4 d \,\operatorname {arctanh}\left (\frac {-\tanh \left (\frac {x \lambda }{2}\right ) b +a}{\sqrt {a^{2}+b^{2}}}\right )}{\lambda \sqrt {a^{2}+b^{2}}}}}{\int \frac {{\mathrm e}^{\frac {c \left (\int \frac {\sinh \left (x \mu \right )}{\sinh \left (x \lambda \right ) a +b}d x \right ) \lambda \sqrt {a^{2}+b^{2}}+4 d \,\operatorname {arctanh}\left (\frac {-\tanh \left (\frac {x \lambda }{2}\right ) b +a}{\sqrt {a^{2}+b^{2}}}\right )}{\lambda \sqrt {a^{2}+b^{2}}}}}{\sinh \left (x \lambda \right ) a +b}d x -c_{1}} \]
✓ Solution by Mathematica
Time used: 28.506 (sec). Leaf size: 289
DSolve[(a*Sinh[\[Lambda]*x]+b)*y'[x]==y[x]^2+c*Sinh[\[Mu]*x]*y[x]-d^2+c*d*Sinh[\[Mu]*x],y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^x-\frac {\exp \left (-\int _1^{K[2]}\frac {2 d-c \sinh (\mu K[1])}{b+a \sinh (\lambda K[1])}dK[1]\right ) (-d+c \sinh (\mu K[2])+y(x))}{c \mu (b+a \sinh (\lambda K[2])) (d+y(x))}dK[2]+\int _1^{y(x)}\left (\frac {\exp \left (-\int _1^x\frac {2 d-c \sinh (\mu K[1])}{b+a \sinh (\lambda K[1])}dK[1]\right )}{c \mu (d+K[3])^2}-\int _1^x\left (\frac {\exp \left (-\int _1^{K[2]}\frac {2 d-c \sinh (\mu K[1])}{b+a \sinh (\lambda K[1])}dK[1]\right ) (-d+K[3]+c \sinh (\mu K[2]))}{c \mu (d+K[3])^2 (b+a \sinh (\lambda K[2]))}-\frac {\exp \left (-\int _1^{K[2]}\frac {2 d-c \sinh (\mu K[1])}{b+a \sinh (\lambda K[1])}dK[1]\right )}{c \mu (d+K[3]) (b+a \sinh (\lambda K[2]))}\right )dK[2]\right )dK[3]=c_1,y(x)\right ] \]