Internal problem ID [9333]
Internal file name [OUTPUT/8269_Monday_June_06_2022_02_35_45_AM_70688720/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 1000.
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]
\[ \boxed {y^{\prime }-\frac {2 y x^{2}+x^{3}+y \ln \left (x \right ) x -y^{2}-x y}{x^{2} \left (x +\ln \left (x \right )\right )}=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {2 y \,x^{2}+x^{3}+x \ln \left (x \right ) y -y^{2}-y x}{x^{2} \left (x +\ln \left (x \right )\right )} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {\ln \left (x \right ) y}{x \left (x +\ln \left (x \right )\right )}+\frac {x}{x +\ln \left (x \right )}+\frac {2 y}{x +\ln \left (x \right )}-\frac {y}{x \left (x +\ln \left (x \right )\right )}-\frac {y^{2}}{x^{2} \left (x +\ln \left (x \right )\right )} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {x}{x +\ln \left (x \right )}\), \(f_1(x)=\frac {2 x^{2}+x \ln \left (x \right )-x}{x^{2} \left (x +\ln \left (x \right )\right )}\) and \(f_2(x)=-\frac {1}{x^{2} \left (x +\ln \left (x \right )\right )}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {u}{x^{2} \left (x +\ln \left (x \right )\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 {2}{x^{3} \left (x +\ln \left (x \right )\right )}+\frac {1+\frac {1}{x}}{x^{2} \left (x +\ln \left (x \right )\right )^{2}}\\ f_1 f_2 &=-\frac {2 x^{2}+x \ln \left (x \right )-x}{x^{4} \left (x +\ln \left (x \right )\right )^{2}}\\ f_2^2 f_0 &=\frac {1}{x^{3} \left (x +\ln \left (x \right )\right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -\frac {u^{\prime \prime }\left (x \right )}{x^{2} \left (x +\ln \left (x \right )\right )}-\left (\frac {2}{x^{3} \left (x +\ln \left (x \right )\right )}+\frac {1+\frac {1}{x}}{x^{2} \left (x +\ln \left (x \right )\right )^{2}}-\frac {2 x^{2}+x \ln \left (x \right )-x}{x^{4} \left (x +\ln \left (x \right )\right )^{2}}\right ) u^{\prime }\left (x \right )+\frac {u \left (x \right )}{x^{3} \left (x +\ln \left (x \right )\right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{-\frac {\left (\int \frac {\ln \left (x \right )+x +2}{x \left (x +\ln \left (x \right )\right )}d x \right )}{2}} \sqrt {x}\, \left (c_{1} +c_{2} \ln \left (x \right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (c_{2} x -c_{1} \right ) {\mathrm e}^{-\frac {\left (\int \frac {\ln \left (x \right )+x +2}{x \left (x +\ln \left (x \right )\right )}d x \right )}{2}}}{\sqrt {x}\, \left (x +\ln \left (x \right )\right )} \] Using the above in (1) gives the solution \[ y = \frac {\left (c_{2} x -c_{1} \right ) x}{c_{1} +c_{2} \ln \left (x \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (x -c_{3} \right ) x}{c_{3} +\ln \left (x \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (x -c_{3} \right ) x}{c_{3} +\ln \left (x \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (x -c_{3} \right ) x}{c_{3} +\ln \left (x \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x^{2} \ln \left (x \right )+y^{\prime } x^{3}-y \ln \left (x \right ) x -2 y x^{2}-x^{3}+y^{2}+x y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 y x^{2}+x^{3}+y \ln \left (x \right ) x -y^{2}-x y}{x^{3}+\ln \left (x \right ) x^{2}} \end {array} \]
Maple trace Kovacic algorithm successful
`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: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -(ln(x)+x+2)*(diff(y(x), x))/(x*(ln(x)+x))+y(x)/((ln(x)+x)^2*x), 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 -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible <- Kovacics algorithm successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 19
dsolve(diff(y(x),x) = 1/x^2*(2*x^2*y(x)+x^3+y(x)*ln(x)*x-y(x)^2-x*y(x))/(x+ln(x)),y(x), singsol=all)
\[ y \left (x \right ) = \frac {x \left (c_{1} x -1\right )}{c_{1} \ln \left (x \right )+1} \]
✓ Solution by Mathematica
Time used: 1.677 (sec). Leaf size: 27
DSolve[y'[x] == (x^3 - x*y[x] + 2*x^2*y[x] + x*Log[x]*y[x] - y[x]^2)/(x^2*(x + Log[x])),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x (x-c_1)}{\log (x)+c_1} \\ y(x)\to -x \\ \end{align*}