9.12 problem 12

9.12.1 Solving as riccati ode
9.12.2 Maple step by step solution

Internal problem ID [10509]
Internal file name [OUTPUT/9457_Monday_June_06_2022_02_38_23_PM_57207150/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 1.2.6-1. Equations with sine
Problem number: 12.
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.

\[ \boxed {\left (a \sin \left (\lambda x \right )+b \right ) y^{\prime }-y^{2}-c \sin \left (\mu x \right ) y=-d^{2}+c d \sin \left (\mu x \right )} \]

9.12.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y^{2}+c \sin \left (\mu x \right ) y -d^{2}+c d \sin \left (\mu x \right )}{a \sin \left (\lambda x \right )+b} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {c d \sin \left (\mu x \right )}{a \sin \left (\lambda x \right )+b}+\frac {c \sin \left (\mu x \right ) y}{a \sin \left (\lambda x \right )+b}-\frac {d^{2}}{a \sin \left (\lambda x \right )+b}+\frac {y^{2}}{a \sin \left (\lambda x \right )+b} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {-d^{2}+c d \sin \left (\mu x \right )}{a \sin \left (\lambda x \right )+b}\), \(f_1(x)=\frac {c \sin \left (\mu x \right )}{a \sin \left (\lambda x \right )+b}\) and \(f_2(x)=\frac {1}{a \sin \left (\lambda x \right )+b}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{a \sin \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 \cos \left (\lambda x \right )}{\left (a \sin \left (\lambda x \right )+b \right )^{2}}\\ f_1 f_2 &=\frac {c \sin \left (\mu x \right )}{\left (a \sin \left (\lambda x \right )+b \right )^{2}}\\ f_2^2 f_0 &=\frac {-d^{2}+c d \sin \left (\mu x \right )}{\left (a \sin \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 \sin \left (\lambda x \right )+b}-\left (-\frac {a \lambda \cos \left (\lambda x \right )}{\left (a \sin \left (\lambda x \right )+b \right )^{2}}+\frac {c \sin \left (\mu x \right )}{\left (a \sin \left (\lambda x \right )+b \right )^{2}}\right ) u^{\prime }\left (x \right )+\frac {\left (-d^{2}+c d \sin \left (\mu x \right )\right ) u \left (x \right )}{\left (a \sin \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.

9.12.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a \sin \left (\lambda x \right )+b \right ) y^{\prime }-y^{2}-c \sin \left (\mu x \right ) y=-d^{2}+c d \sin \left (\mu x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2}+c \sin \left (\mu x \right ) y-d^{2}+c d \sin \left (\mu x \right )}{a \sin \left (\lambda x \right )+b} \end {array} \]

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: 0.015 (sec). Leaf size: 265

dsolve((a*sin(lambda*x)+b)*diff(y(x),x)=y(x)^2+c*sin(mu*x)*y(x)-d^2+c*d*sin(mu*x),y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {-d \left (\int \frac {{\mathrm e}^{\frac {c \left (\int \frac {\sin \left (x \mu \right )}{a \sin \left (x \lambda \right )+b}d x \right ) \sqrt {-a^{2}+b^{2}}\, \lambda -4 d \arctan \left (\frac {\tan \left (\frac {x \lambda }{2}\right ) b +a}{\sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \lambda }}}{a \sin \left (x \lambda \right )+b}d x \right )+d c_{1} -{\mathrm e}^{\frac {c \left (\int \frac {\sin \left (x \mu \right )}{a \sin \left (x \lambda \right )+b}d x \right ) \sqrt {-a^{2}+b^{2}}\, \lambda -4 d \arctan \left (\frac {\tan \left (\frac {x \lambda }{2}\right ) b +a}{\sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \lambda }}}{\int \frac {{\mathrm e}^{\frac {c \left (\int \frac {\sin \left (x \mu \right )}{a \sin \left (x \lambda \right )+b}d x \right ) \sqrt {-a^{2}+b^{2}}\, \lambda -4 d \arctan \left (\frac {\tan \left (\frac {x \lambda }{2}\right ) b +a}{\sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \lambda }}}{a \sin \left (x \lambda \right )+b}d x -c_{1}} \]

Solution by Mathematica

Time used: 15.846 (sec). Leaf size: 289

DSolve[(a*Sin[\[Lambda]*x]+b)*y'[x]==y[x]^2+c*Sin[\[Mu]*x]*y[x]-d^2+c*d*Sin[\[Mu]*x],y[x],x,IncludeSingularSolutions -> True]
 

\[ \text {Solve}\left [\int _1^x-\frac {\exp \left (-\int _1^{K[2]}\frac {2 d-c \sin (\mu K[1])}{b+a \sin (\lambda K[1])}dK[1]\right ) (-d+c \sin (\mu K[2])+y(x))}{c \mu (b+a \sin (\lambda K[2])) (d+y(x))}dK[2]+\int _1^{y(x)}\left (\frac {\exp \left (-\int _1^x\frac {2 d-c \sin (\mu K[1])}{b+a \sin (\lambda K[1])}dK[1]\right )}{c \mu (d+K[3])^2}-\int _1^x\left (\frac {\exp \left (-\int _1^{K[2]}\frac {2 d-c \sin (\mu K[1])}{b+a \sin (\lambda K[1])}dK[1]\right ) (-d+K[3]+c \sin (\mu K[2]))}{c \mu (d+K[3])^2 (b+a \sin (\lambda K[2]))}-\frac {\exp \left (-\int _1^{K[2]}\frac {2 d-c \sin (\mu K[1])}{b+a \sin (\lambda K[1])}dK[1]\right )}{c \mu (d+K[3]) (b+a \sin (\lambda K[2]))}\right )dK[2]\right )dK[3]=c_1,y(x)\right ] \]