Internal problem ID [9326]
Internal file name [OUTPUT/8262_Monday_June_06_2022_02_34_37_AM_5966859/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 993.
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]
\[ \boxed {y^{\prime }+F \left (x \right ) \left (-y^{2}-2 y \ln \left (x \right )-\ln \left (x \right )^{2}\right )-\frac {y}{\ln \left (x \right ) x}=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {F \left (x \right ) \ln \left (x \right )^{3} x +2 F \left (x \right ) \ln \left (x \right )^{2} y x +F \left (x \right ) y^{2} \ln \left (x \right ) x +y}{\ln \left (x \right ) x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = F \left (x \right ) \ln \left (x \right )^{2}+2 F \left (x \right ) \ln \left (x \right ) y +F \left (x \right ) y^{2}+\frac {y}{\ln \left (x \right ) x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=F \left (x \right ) \ln \left (x \right )^{2}\), \(f_1(x)=\frac {2 \ln \left (x \right )^{2} F \left (x \right ) x +1}{\ln \left (x \right ) x}\) and \(f_2(x)=F \left (x \right )\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{F \left (x \right ) 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' &=F^{\prime }\left (x \right )\\ f_1 f_2 &=\frac {\left (2 \ln \left (x \right )^{2} F \left (x \right ) x +1\right ) F \left (x \right )}{\ln \left (x \right ) x}\\ f_2^2 f_0 &=F \left (x \right )^{3} \ln \left (x \right )^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} F \left (x \right ) u^{\prime \prime }\left (x \right )-\left (F^{\prime }\left (x \right )+\frac {\left (2 \ln \left (x \right )^{2} F \left (x \right ) x +1\right ) F \left (x \right )}{\ln \left (x \right ) x}\right ) u^{\prime }\left (x \right )+F \left (x \right )^{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
\[ u \left (x \right ) = {\mathrm e}^{\int F \left (x \right ) \ln \left (x \right )d x} \left (c_{1} -i \left (\int F \left (x \right ) \ln \left (x \right )d x \right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = F \left (x \right ) \ln \left (x \right ) {\mathrm e}^{\int F \left (x \right ) \ln \left (x \right )d x} \left (-i \left (\int F \left (x \right ) \ln \left (x \right )d x \right ) c_{2} -i c_{2} +c_{1} \right ) \] Using the above in (1) gives the solution \[ y = -\frac {\ln \left (x \right ) \left (-i \left (\int F \left (x \right ) \ln \left (x \right )d x \right ) c_{2} -i c_{2} +c_{1} \right )}{c_{1} -i \left (\int F \left (x \right ) \ln \left (x \right )d x \right ) c_{2}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\ln \left (x \right ) \left (\int F \left (x \right ) \ln \left (x \right )d x +1+i c_{3} \right )}{\int F \left (x \right ) \ln \left (x \right )d x +i c_{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\ln \left (x \right ) \left (\int F \left (x \right ) \ln \left (x \right )d x +1+i c_{3} \right )}{\int F \left (x \right ) \ln \left (x \right )d x +i c_{3}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\ln \left (x \right ) \left (\int F \left (x \right ) \ln \left (x \right )d x +1+i c_{3} \right )}{\int F \left (x \right ) \ln \left (x \right )d x +i c_{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & F \left (x \right ) y^{2} \ln \left (x \right ) x +2 F \left (x \right ) y \ln \left (x \right )^{2} x +F \left (x \right ) \ln \left (x \right )^{3} x -y^{\prime } \ln \left (x \right ) x +y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {-F \left (x \right ) \ln \left (x \right )^{3} x -2 F \left (x \right ) y \ln \left (x \right )^{2} x -F \left (x \right ) y^{2} \ln \left (x \right ) x -y}{x \ln \left (x \right )} \end {array} \]
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: 33
dsolve(diff(y(x),x) = -F(x)*(-y(x)^2-2*y(x)*ln(x)-ln(x)^2)+1/ln(x)/x*y(x),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\ln \left (x \right ) \left (2 \left (\int \ln \left (x \right ) F \left (x \right )d x \right )+c_{1} +2\right )}{2 \left (\int \ln \left (x \right ) F \left (x \right )d x \right )+c_{1}} \]
✓ Solution by Mathematica
Time used: 3.065 (sec). Leaf size: 75
DSolve[y'[x] == y[x]/(x*Log[x]) - F[x]*(-Log[x]^2 - 2*Log[x]*y[x] - y[x]^2),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\int _1^x\frac {F(K[1])}{\sqrt {\frac {1}{\log ^2(K[1])}}}dK[1]-1+c_1}{\sqrt {\frac {1}{\log ^2(x)}} \left (\int _1^x\frac {F(K[1])}{\sqrt {\frac {1}{\log ^2(K[1])}}}dK[1]+c_1\right )} \\ y(x)\to \frac {1}{\sqrt {\frac {1}{\log ^2(x)}}} \\ \end{align*}