Internal problem ID [8530]
Internal file name [OUTPUT/7463_Sunday_June_05_2022_10_55_37_PM_39022776/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 194.
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]
\[ \boxed {x y^{\prime } \ln \left (x \right )-y^{2} \ln \left (x \right )-\left (2 \ln \left (x \right )^{2}+1\right ) y=\ln \left (x \right )^{3}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {\ln \left (x \right )^{3}+2 \ln \left (x \right )^{2} y +y^{2} \ln \left (x \right )+y}{x \ln \left (x \right )} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {\ln \left (x \right )^{2}}{x}+\frac {2 \ln \left (x \right ) y}{x}+\frac {y^{2}}{x}+\frac {y}{x \ln \left (x \right )} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {\ln \left (x \right )^{2}}{x}\), \(f_1(x)=\frac {2 \ln \left (x \right )^{2}+1}{x \ln \left (x \right )}\) 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 \ln \left (x \right )^{2}+1}{x^{2} \ln \left (x \right )}\\ f_2^2 f_0 &=\frac {\ln \left (x \right )^{2}}{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 \ln \left (x \right )^{2}+1}{x^{2} \ln \left (x \right )}\right ) u^{\prime }\left (x \right )+\frac {\ln \left (x \right )^{2} u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\frac {\ln \left (x \right )^{2}}{2}} \left (c_{1} +\ln \left (x \right )^{2} c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\ln \left (x \right ) {\mathrm e}^{\frac {\ln \left (x \right )^{2}}{2}} \left (\ln \left (x \right )^{2} c_{2} +c_{1} +2 c_{2} \right )}{x} \] Using the above in (1) gives the solution \[ y = -\frac {\ln \left (x \right ) \left (\ln \left (x \right )^{2} c_{2} +c_{1} +2 c_{2} \right )}{c_{1} +\ln \left (x \right )^{2} c_{2}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\ln \left (x \right ) \left (\ln \left (x \right )^{2}+c_{3} +2\right )}{c_{3} +\ln \left (x \right )^{2}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\ln \left (x \right ) \left (\ln \left (x \right )^{2}+c_{3} +2\right )}{c_{3} +\ln \left (x \right )^{2}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\ln \left (x \right ) \left (\ln \left (x \right )^{2}+c_{3} +2\right )}{c_{3} +\ln \left (x \right )^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime } \ln \left (x \right )-y^{2} \ln \left (x \right )-\left (2 \ln \left (x \right )^{2}+1\right ) y=\ln \left (x \right )^{3} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2} \ln \left (x \right )+\left (2 \ln \left (x \right )^{2}+1\right ) y+\ln \left (x \right )^{3}}{x \ln \left (x \right )} \end {array} \]
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: 23
dsolve(x*diff(y(x),x)*ln(x) - y(x)^2*ln(x) - (2*ln(x)^2+1)*y(x) - ln(x)^3=0,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\ln \left (x \right ) \left (\ln \left (x \right )^{2}+c_{1} +2\right )}{\ln \left (x \right )^{2}+c_{1}} \]
✓ Solution by Mathematica
Time used: 0.332 (sec). Leaf size: 38
DSolve[x*y'[x]*Log[x] - y[x]^2*Log[x] - (2*Log[x]^2+1)*y[x] - Log[x]^3==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {\log (x) \left (\log ^2(x)+2+2 c_1\right )}{\log ^2(x)+2 c_1} \\ y(x)\to -\log (x) \\ \end{align*}