Internal problem ID [9177]
Internal file name [OUTPUT/8112_Monday_June_06_2022_01_47_46_AM_22175166/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 843.
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]
\[ \boxed {y^{\prime }-\frac {y+\ln \left (x \right )^{3} x^{3}+2 x^{3} \ln \left (x \right )^{2} y+y^{2} x^{3} \ln \left (x \right )}{x \ln \left (x \right )}=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y +\ln \left (x \right )^{3} x^{3}+2 \ln \left (x \right )^{2} x^{3} y +\ln \left (x \right ) x^{3} y^{2}}{x \ln \left (x \right )} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \ln \left (x \right )^{2} x^{2}+2 y \,x^{2} \ln \left (x \right )+x^{2} y^{2}+\frac {y}{x \ln \left (x \right )} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\ln \left (x \right )^{2} x^{2}\), \(f_1(x)=\frac {2 \ln \left (x \right )^{2} x^{3}+1}{x \ln \left (x \right )}\) and \(f_2(x)=x^{2}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{x^{2} 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' &=2 x\\ f_1 f_2 &=\frac {\left (2 \ln \left (x \right )^{2} x^{3}+1\right ) x}{\ln \left (x \right )}\\ f_2^2 f_0 &=\ln \left (x \right )^{2} x^{6} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} x^{2} u^{\prime \prime }\left (x \right )-\left (2 x +\frac {\left (2 \ln \left (x \right )^{2} x^{3}+1\right ) x}{\ln \left (x \right )}\right ) u^{\prime }\left (x \right )+\ln \left (x \right )^{2} x^{6} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{\frac {x^{3}}{3}} {\mathrm e}^{-\frac {x^{3}}{9}} \left (3 \ln \left (x \right ) x^{3} c_{2} -c_{2} x^{3}+c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = x^{2+\frac {x^{3}}{3}} \ln \left (x \right ) {\mathrm e}^{-\frac {x^{3}}{9}} \left (3 \ln \left (x \right ) x^{3} c_{2} -c_{2} x^{3}+c_{1} +9 c_{2} \right ) \] Using the above in (1) gives the solution \[ y = -\frac {x^{2+\frac {x^{3}}{3}} \ln \left (x \right ) \left (3 \ln \left (x \right ) x^{3} c_{2} -c_{2} x^{3}+c_{1} +9 c_{2} \right ) x^{-\frac {x^{3}}{3}}}{x^{2} \left (3 \ln \left (x \right ) x^{3} c_{2} -c_{2} x^{3}+c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\ln \left (x \right ) \left (3 x^{3} \ln \left (x \right )-x^{3}+c_{3} +9\right )}{3 x^{3} \ln \left (x \right )-x^{3}+c_{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\ln \left (x \right ) \left (3 x^{3} \ln \left (x \right )-x^{3}+c_{3} +9\right )}{3 x^{3} \ln \left (x \right )-x^{3}+c_{3}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\ln \left (x \right ) \left (3 x^{3} \ln \left (x \right )-x^{3}+c_{3} +9\right )}{3 x^{3} \ln \left (x \right )-x^{3}+c_{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \ln \left (x \right )^{3} x^{3}+2 x^{3} \ln \left (x \right )^{2} y+y^{2} x^{3} \ln \left (x \right )-y^{\prime } x \ln \left (x \right )+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {-y-\ln \left (x \right )^{3} x^{3}-2 x^{3} \ln \left (x \right )^{2} y-y^{2} x^{3} \ln \left (x \right )}{x \ln \left (x \right )} \end {array} \]
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 Chini <- Chini successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 43
dsolve(diff(y(x),x) = (y(x)+x^3*ln(x)^3+2*x^3*ln(x)^2*y(x)+x^3*ln(x)*y(x)^2)/x/ln(x),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\ln \left (x \right ) \left (6 x^{3} \ln \left (x \right )-2 x^{3}+9 c_{1} +18\right )}{6 x^{3} \ln \left (x \right )-2 x^{3}+9 c_{1}} \]
✓ Solution by Mathematica
Time used: 0.378 (sec). Leaf size: 52
DSolve[y'[x] == (x^3*Log[x]^3 + y[x] + 2*x^3*Log[x]^2*y[x] + x^3*Log[x]*y[x]^2)/(x*Log[x]),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\log (x) \left (x^3-3 x^3 \log (x)-9 (1+c_1)\right )}{-x^3+3 x^3 \log (x)+9 c_1} \\ y(x)\to -\log (x) \\ \end{align*}