2.210 problem 786

2.210.1 Solving as homogeneousTypeD2 ode
2.210.2 Solving as riccati ode
2.210.3 Maple step by step solution

Internal problem ID [9120]
Internal file name [OUTPUT/8055_Monday_June_06_2022_01_27_13_AM_43111602/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 786.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "riccati", "homogeneousTypeD2"

Maple gives the following as the ode type

[[_homogeneous, `class D`], _Riccati]

\[ \boxed {y^{\prime }-\frac {y \ln \left (x \right )+\cosh \left (x \right ) x a y^{2}+\cosh \left (x \right ) x^{3} b}{x \ln \left (x \right )}=0} \]

2.210.1 Solving as homogeneousTypeD2 ode

Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} \cosh \left (x \right ) x^{3} a u \left (x \right )^{2}+\cosh \left (x \right ) x^{3} b -\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x \ln \left (x \right )+u \left (x \right ) x \ln \left (x \right ) = 0 \end {align*}

In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {\cosh \left (x \right ) x \left (u^{2} a +b \right )}{\ln \left (x \right )} \end {align*}

Where \(f(x)=\frac {\cosh \left (x \right ) x}{\ln \left (x \right )}\) and \(g(u)=u^{2} a +b\). Integrating both sides gives \begin{align*} \frac {1}{u^{2} a +b} \,du &= \frac {\cosh \left (x \right ) x}{\ln \left (x \right )} \,d x \\ \int { \frac {1}{u^{2} a +b} \,du} &= \int {\frac {\cosh \left (x \right ) x}{\ln \left (x \right )} \,d x} \\ \frac {\arctan \left (\frac {a u}{\sqrt {a b}}\right )}{\sqrt {a b}}&=\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x +c_{2} \\ \end{align*} The solution is \[ \frac {\arctan \left (\frac {a u \left (x \right )}{\sqrt {a b}}\right )}{\sqrt {a b}}-\left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {\arctan \left (\frac {y a}{x \sqrt {a b}}\right )}{\sqrt {a b}}-\left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )-c_{2} = 0\\ \frac {\arctan \left (\frac {y a}{x \sqrt {a b}}\right )}{\sqrt {a b}}-\left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )-c_{2} = 0 \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {\arctan \left (\frac {y a}{x \sqrt {a b}}\right )}{\sqrt {a b}}-\left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )-c_{2} &= 0 \\ \end{align*}

Verification of solutions

\[ \frac {\arctan \left (\frac {y a}{x \sqrt {a b}}\right )}{\sqrt {a b}}-\left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )-c_{2} = 0 \] Verified OK.

2.210.2 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y \ln \left (x \right )+\cosh \left (x \right ) x a \,y^{2}+\cosh \left (x \right ) x^{3} b}{x \ln \left (x \right )} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {\cosh \left (x \right ) a \,y^{2}}{\ln \left (x \right )}+\frac {x^{2} \cosh \left (x \right ) b}{\ln \left (x \right )}+\frac {y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {x^{2} \cosh \left (x \right ) b}{\ln \left (x \right )}\), \(f_1(x)=\frac {1}{x}\) and \(f_2(x)=\frac {a \cosh \left (x \right )}{\ln \left (x \right )}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {a \cosh \left (x \right ) u}{\ln \left (x \right )}} \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 \cosh \left (x \right )}{\ln \left (x \right )^{2} x}+\frac {a \sinh \left (x \right )}{\ln \left (x \right )}\\ f_1 f_2 &=\frac {a \cosh \left (x \right )}{x \ln \left (x \right )}\\ f_2^2 f_0 &=\frac {a^{2} \cosh \left (x \right )^{3} x^{2} b}{\ln \left (x \right )^{3}} \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} \frac {a \cosh \left (x \right ) u^{\prime \prime }\left (x \right )}{\ln \left (x \right )}-\left (-\frac {a \cosh \left (x \right )}{\ln \left (x \right )^{2} x}+\frac {a \sinh \left (x \right )}{\ln \left (x \right )}+\frac {a \cosh \left (x \right )}{x \ln \left (x \right )}\right ) u^{\prime }\left (x \right )+\frac {a^{2} \cosh \left (x \right )^{3} x^{2} b u \left (x \right )}{\ln \left (x \right )^{3}} &=0 \end {align*}

Solving the above ODE (this ode solved using Maple, not this program), gives

\[ u \left (x \right ) = c_{1} {\mathrm e}^{i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}+c_{2} {\mathrm e}^{-i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {i \cosh \left (x \right ) x \sqrt {a b}\, \left (c_{1} {\mathrm e}^{i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}-c_{2} {\mathrm e}^{-i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}\right )}{\ln \left (x \right )} \] Using the above in (1) gives the solution \[ y = -\frac {i x \sqrt {a b}\, \left (c_{1} {\mathrm e}^{i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}-c_{2} {\mathrm e}^{-i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}\right )}{a \left (c_{1} {\mathrm e}^{i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}+c_{2} {\mathrm e}^{-i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = -\frac {i x \sqrt {a b}\, \left (c_{3} {\mathrm e}^{i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}-{\mathrm e}^{-i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}\right )}{a \left (c_{3} {\mathrm e}^{i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}+{\mathrm e}^{-i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {i x \sqrt {a b}\, \left (c_{3} {\mathrm e}^{i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}-{\mathrm e}^{-i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}\right )}{a \left (c_{3} {\mathrm e}^{i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}+{\mathrm e}^{-i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}\right )} \\ \end{align*}

Verification of solutions

\[ y = -\frac {i x \sqrt {a b}\, \left (c_{3} {\mathrm e}^{i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}-{\mathrm e}^{-i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}\right )}{a \left (c_{3} {\mathrm e}^{i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}+{\mathrm e}^{-i \sqrt {a b}\, \left (\int \frac {\cosh \left (x \right ) x}{\ln \left (x \right )}d x \right )}\right )} \] Verified OK.

2.210.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \cosh \left (x \right ) x a y^{2}+\cosh \left (x \right ) x^{3} b -y^{\prime } x \ln \left (x \right )+y \ln \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y \ln \left (x \right )+\cosh \left (x \right ) x a y^{2}+\cosh \left (x \right ) x^{3} b}{x \ln \left (x \right )} \end {array} \]

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 homogeneous D 
<- homogeneous successful`
 

Solution by Maple

Time used: 0.047 (sec). Leaf size: 33

dsolve(diff(y(x),x) = (y(x)*ln(x)+cosh(x)*x*a*y(x)^2+cosh(x)*x^3*b)/x/ln(x),y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\tan \left (\sqrt {a b}\, \left (\int \frac {x \cosh \left (x \right )}{\ln \left (x \right )}d x +c_{1} \right )\right ) x \sqrt {a b}}{a} \]

Solution by Mathematica

Time used: 6.061 (sec). Leaf size: 50

DSolve[y'[x] == (b*x^3*Cosh[x] + Log[x]*y[x] + a*x*Cosh[x]*y[x]^2)/(x*Log[x]),y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {\sqrt {b} x \tan \left (\sqrt {a} \sqrt {b} \left (\int _1^x\frac {\cosh (K[1]) K[1]}{\log (K[1])}dK[1]+c_1\right )\right )}{\sqrt {a}} \]