Internal problem ID [9204]
Internal file name [OUTPUT/8140_Monday_June_06_2022_01_53_45_AM_15454070/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 871.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]
\[ \boxed {y^{\prime }-\frac {2 y^{2} x +4 y \ln \left (2 x +1\right ) x +2 \ln \left (2 x +1\right )^{2} x +y^{2}-2+\ln \left (2 x +1\right )^{2}+2 y \ln \left (2 x +1\right )}{2 x +1}=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {2 x \,y^{2}+4 y \ln \left (2 x +1\right ) x +2 \ln \left (2 x +1\right )^{2} x +y^{2}-2+\ln \left (2 x +1\right )^{2}+2 y \ln \left (2 x +1\right )}{2 x +1} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {2 \ln \left (2 x +1\right )^{2} x}{2 x +1}+\frac {4 y \ln \left (2 x +1\right ) x}{2 x +1}+\frac {2 x \,y^{2}}{2 x +1}+\frac {\ln \left (2 x +1\right )^{2}}{2 x +1}+\frac {2 y \ln \left (2 x +1\right )}{2 x +1}+\frac {y^{2}}{2 x +1}-\frac {2}{2 x +1} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {2 \ln \left (2 x +1\right )^{2} x +\ln \left (2 x +1\right )^{2}-2}{2 x +1}\), \(f_1(x)=\frac {4 \ln \left (2 x +1\right ) x +2 \ln \left (2 x +1\right )}{2 x +1}\) 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 &=\frac {4 \ln \left (2 x +1\right ) x +2 \ln \left (2 x +1\right )}{2 x +1}\\ f_2^2 f_0 &=\frac {2 \ln \left (2 x +1\right )^{2} x +\ln \left (2 x +1\right )^{2}-2}{2 x +1} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )-\frac {\left (4 \ln \left (2 x +1\right ) x +2 \ln \left (2 x +1\right )\right ) u^{\prime }\left (x \right )}{2 x +1}+\frac {\left (2 \ln \left (2 x +1\right )^{2} x +\ln \left (2 x +1\right )^{2}-2\right ) u \left (x \right )}{2 x +1} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (2 x +1\right )^{x +\frac {1}{2}} {\mathrm e}^{-x} \left (c_{2} x +c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \left (2 x +1\right )^{x +\frac {1}{2}} {\mathrm e}^{-x} \left (\left (c_{2} x +c_{1} \right ) \ln \left (2 x +1\right )+c_{2} \right ) \] Using the above in (1) gives the solution \[ y = -\frac {\left (c_{2} x +c_{1} \right ) \ln \left (2 x +1\right )+c_{2}}{c_{2} x +c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-1+\left (-x -c_{3} \right ) \ln \left (2 x +1\right )}{c_{3} +x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-1+\left (-x -c_{3} \right ) \ln \left (2 x +1\right )}{c_{3} +x} \\ \end{align*}
Verification of solutions
\[ y = \frac {-1+\left (-x -c_{3} \right ) \ln \left (2 x +1\right )}{c_{3} +x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -2 y^{2} x -4 y \ln \left (2 x +1\right ) x -2 \ln \left (2 x +1\right )^{2} x +2 y^{\prime } x -y^{2}-2 y \ln \left (2 x +1\right )-\ln \left (2 x +1\right )^{2}+y^{\prime }+2=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 y^{2} x +4 y \ln \left (2 x +1\right ) x +2 \ln \left (2 x +1\right )^{2} x +y^{2}-2+\ln \left (2 x +1\right )^{2}+2 y \ln \left (2 x +1\right )}{2 x +1} \end {array} \]
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 (d) successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 26
dsolve(diff(y(x),x) = 1/(2*x+1)*(2*x*y(x)^2+4*y(x)*ln(2*x+1)*x+2*ln(2*x+1)^2*x+y(x)^2-2+ln(2*x+1)^2+2*y(x)*ln(2*x+1)),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-1+\left (-x +c_{1} \right ) \ln \left (2 x +1\right )}{x -c_{1}} \]
✓ Solution by Mathematica
Time used: 0.351 (sec). Leaf size: 34
DSolve[y'[x] == (-2 + Log[1 + 2*x]^2 + 2*x*Log[1 + 2*x]^2 + 2*Log[1 + 2*x]*y[x] + 4*x*Log[1 + 2*x]*y[x] + y[x]^2 + 2*x*y[x]^2)/(1 + 2*x),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\log (2 x+1)+\frac {1}{-x+c_1} \\ y(x)\to -\log (2 x+1) \\ \end{align*}