Internal problem ID [9176]
Internal file name [OUTPUT/8111_Monday_June_06_2022_01_47_36_AM_44707634/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 842.
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]
\[ \boxed {y^{\prime }-\frac {y+x^{2} \ln \left (x \right )^{3}+2 x^{2} \ln \left (x \right )^{2} y+x^{2} \ln \left (x \right ) y^{2}}{x \ln \left (x \right )}=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y +x^{2} \ln \left (x \right )^{3}+2 x^{2} \ln \left (x \right )^{2} y +x^{2} \ln \left (x \right ) y^{2}}{x \ln \left (x \right )} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = x \ln \left (x \right )^{2}+2 x \ln \left (x \right ) y +x \,y^{2}+\frac {y}{x \ln \left (x \right )} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=x \ln \left (x \right )^{2}\), \(f_1(x)=\frac {2 \ln \left (x \right )^{2} x^{2}+1}{x \ln \left (x \right )}\) and \(f_2(x)=x\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{x 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' &=1\\ f_1 f_2 &=\frac {2 \ln \left (x \right )^{2} x^{2}+1}{\ln \left (x \right )}\\ f_2^2 f_0 &=x^{3} \ln \left (x \right )^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} x u^{\prime \prime }\left (x \right )-\left (1+\frac {2 \ln \left (x \right )^{2} x^{2}+1}{\ln \left (x \right )}\right ) u^{\prime }\left (x \right )+x^{3} \ln \left (x \right )^{2} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{\frac {x^{2}}{2}} {\mathrm e}^{-\frac {x^{2}}{4}} \left (2 c_{2} x^{2} \ln \left (x \right )-c_{2} x^{2}+c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = x^{1+\frac {x^{2}}{2}} \ln \left (x \right ) {\mathrm e}^{-\frac {x^{2}}{4}} \left (2 c_{2} x^{2} \ln \left (x \right )-c_{2} x^{2}+c_{1} +4 c_{2} \right ) \] Using the above in (1) gives the solution \[ y = -\frac {x^{1+\frac {x^{2}}{2}} \ln \left (x \right ) \left (2 c_{2} x^{2} \ln \left (x \right )-c_{2} x^{2}+c_{1} +4 c_{2} \right ) x^{-\frac {x^{2}}{2}}}{x \left (2 c_{2} x^{2} \ln \left (x \right )-c_{2} x^{2}+c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\ln \left (x \right ) \left (2 \ln \left (x \right ) x^{2}-x^{2}+c_{3} +4\right )}{2 \ln \left (x \right ) x^{2}-x^{2}+c_{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\ln \left (x \right ) \left (2 \ln \left (x \right ) x^{2}-x^{2}+c_{3} +4\right )}{2 \ln \left (x \right ) x^{2}-x^{2}+c_{3}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\ln \left (x \right ) \left (2 \ln \left (x \right ) x^{2}-x^{2}+c_{3} +4\right )}{2 \ln \left (x \right ) x^{2}-x^{2}+c_{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \ln \left (x \right )^{3}+2 x^{2} \ln \left (x \right )^{2} y+x^{2} \ln \left (x \right ) y^{2}-y^{\prime } x \ln \left (x \right )+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {-y-x^{2} \ln \left (x \right )^{3}-2 x^{2} \ln \left (x \right )^{2} y-x^{2} \ln \left (x \right ) y^{2}}{x \ln \left (x \right )} \end {array} \]
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: 0.016 (sec). Leaf size: 43
dsolve(diff(y(x),x) = (y(x)+x^2*ln(x)^3+2*x^2*ln(x)^2*y(x)+x^2*ln(x)*y(x)^2)/x/ln(x),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\ln \left (x \right ) \left (2 x^{2} \ln \left (x \right )-x^{2}+2 c_{1} +4\right )}{2 x^{2} \ln \left (x \right )-x^{2}+2 c_{1}} \]
✓ Solution by Mathematica
Time used: 0.373 (sec). Leaf size: 52
DSolve[y'[x] == (x^2*Log[x]^3 + y[x] + 2*x^2*Log[x]^2*y[x] + x^2*Log[x]*y[x]^2)/(x*Log[x]),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\log (x) \left (x^2-2 x^2 \log (x)-4 (1+c_1)\right )}{-x^2+2 x^2 \log (x)+4 c_1} \\ y(x)\to -\log (x) \\ \end{align*}