Internal problem ID [9144]
Internal file name [OUTPUT/8079_Monday_June_06_2022_01_41_09_AM_63277471/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 810.
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]
\[ \boxed {y^{\prime }-\frac {x +y+y^{2}-2 y \ln \left (x \right ) x +\ln \left (x \right )^{2} x^{2}}{x}=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x +y +y^{2}-2 x y \ln \left (x \right )+\ln \left (x \right )^{2} x^{2}}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \ln \left (x \right )^{2} x -2 y \ln \left (x \right )+\frac {y^{2}}{x}+1+\frac {y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {\ln \left (x \right )^{2} x^{2}+x}{x}\), \(f_1(x)=\frac {-2 x \ln \left (x \right )+1}{x}\) and \(f_2(x)=\frac {1}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{x}} \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 {1}{x^{2}}\\ f_1 f_2 &=\frac {-2 x \ln \left (x \right )+1}{x^{2}}\\ f_2^2 f_0 &=\frac {\ln \left (x \right )^{2} x^{2}+x}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {u^{\prime \prime }\left (x \right )}{x}-\left (-\frac {1}{x^{2}}+\frac {-2 x \ln \left (x \right )+1}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {\left (\ln \left (x \right )^{2} x^{2}+x \right ) u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{-x} {\mathrm e}^{x} \left (c_{2} x +c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\left (\left (c_{2} x +c_{1} \right ) \ln \left (x \right )-c_{2} \right ) x^{-x} {\mathrm e}^{x} \] Using the above in (1) gives the solution \[ y = \frac {\left (\left (c_{2} x +c_{1} \right ) \ln \left (x \right )-c_{2} \right ) x}{c_{2} x +c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (\left (c_{3} +x \right ) \ln \left (x \right )-1\right ) x}{c_{3} +x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\left (c_{3} +x \right ) \ln \left (x \right )-1\right ) x}{c_{3} +x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (\left (c_{3} +x \right ) \ln \left (x \right )-1\right ) x}{c_{3} +x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \ln \left (x \right )^{2} x^{2}-2 y \ln \left (x \right ) x +y^{2}-y^{\prime } x +y+x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {-x -y-y^{2}+2 y \ln \left (x \right ) x -\ln \left (x \right )^{2} x^{2}}{x} \end {array} \]
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 polynomial solution successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 16
dsolve(diff(y(x),x) = (x+y(x)+y(x)^2-2*y(x)*ln(x)*x+x^2*ln(x)^2)/x,y(x), singsol=all)
\[ y \left (x \right ) = \left (\ln \left (x \right )+\frac {1}{-x +c_{1}}\right ) x \]
✓ Solution by Mathematica
Time used: 0.21 (sec). Leaf size: 26
DSolve[y'[x] == (x + x^2*Log[x]^2 + y[x] - 2*x*Log[x]*y[x] + y[x]^2)/x,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to x \left (\log (x)+\frac {1}{-x+c_1}\right ) \\ y(x)\to x \log (x) \\ \end{align*}