Internal problem ID [9029]
Internal file name [OUTPUT/7964_Monday_June_06_2022_01_02_14_AM_22402907/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 695.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati", "homogeneousTypeD2"
Maple gives the following as the ode type
[[_homogeneous, `class D`], _Riccati]
\[ \boxed {y^{\prime }-\frac {y \ln \left (x -1\right )+x^{4}+x^{3}+y^{2} x^{2}+y^{2} x}{\ln \left (x -1\right ) x}=0} \]
Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} -u \left (x \right )^{2} x^{4}-x^{4}+\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) \ln \left (x -1\right ) x -u \left (x \right )^{2} x^{3}-x^{3}-u \left (x \right ) x \ln \left (x -1\right ) = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {x \left (u^{2}+1\right ) \left (x +1\right )}{\ln \left (x -1\right )} \end {align*}
Where \(f(x)=\frac {x \left (x +1\right )}{\ln \left (x -1\right )}\) and \(g(u)=u^{2}+1\). Integrating both sides gives \begin{align*} \frac {1}{u^{2}+1} \,du &= \frac {x \left (x +1\right )}{\ln \left (x -1\right )} \,d x \\ \int { \frac {1}{u^{2}+1} \,du} &= \int {\frac {x \left (x +1\right )}{\ln \left (x -1\right )} \,d x} \\ \arctan \left (u \right )&=-\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )-3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )-2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )+c_{2} \\ \end{align*} The solution is \[ \arctan \left (u \left (x \right )\right )+\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \arctan \left (\frac {y}{x}\right )+\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )-c_{2} = 0\\ \arctan \left (\frac {y}{x}\right )+\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )-c_{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \arctan \left (\frac {y}{x}\right )+\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \arctan \left (\frac {y}{x}\right )+\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )-c_{2} = 0 \] Verified OK.
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y \ln \left (x -1\right )+x^{4}+x^{3}+x^{2} y^{2}+x \,y^{2}}{\ln \left (x -1\right ) x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {x^{3}}{\ln \left (x -1\right )}+\frac {x \,y^{2}}{\ln \left (x -1\right )}+\frac {x^{2}}{\ln \left (x -1\right )}+\frac {y^{2}}{\ln \left (x -1\right )}+\frac {y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {x^{4}+x^{3}}{\ln \left (x -1\right ) x}\), \(f_1(x)=\frac {1}{x}\) and \(f_2(x)=\frac {x^{2}+x}{\ln \left (x -1\right ) x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {\left (x^{2}+x \right ) u}{\ln \left (x -1\right ) 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 {2 x +1}{\ln \left (x -1\right ) x}-\frac {x^{2}+x}{\ln \left (x -1\right )^{2} x \left (x -1\right )}-\frac {x^{2}+x}{\ln \left (x -1\right ) x^{2}}\\ f_1 f_2 &=\frac {x^{2}+x}{\ln \left (x -1\right ) x^{2}}\\ f_2^2 f_0 &=\frac {\left (x^{2}+x \right )^{2} \left (x^{4}+x^{3}\right )}{\ln \left (x -1\right )^{3} x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {\left (x^{2}+x \right ) u^{\prime \prime }\left (x \right )}{\ln \left (x -1\right ) x}-\left (\frac {2 x +1}{\ln \left (x -1\right ) x}-\frac {x^{2}+x}{\ln \left (x -1\right )^{2} x \left (x -1\right )}\right ) u^{\prime }\left (x \right )+\frac {\left (x^{2}+x \right )^{2} \left (x^{4}+x^{3}\right ) u \left (x \right )}{\ln \left (x -1\right )^{3} x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \sin \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )+c_{2} \cos \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (-c_{1} \cos \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )+c_{2} \sin \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )\right ) x \left (x +1\right )}{\ln \left (x -1\right )} \] Using the above in (1) gives the solution \[ y = -\frac {\left (-c_{1} \cos \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )+c_{2} \sin \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )\right ) x^{2} \left (x +1\right )}{\left (x^{2}+x \right ) \left (c_{1} \sin \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )+c_{2} \cos \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (c_{3} \cos \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )-\sin \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )\right ) x}{c_{3} \sin \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )+\cos \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (c_{3} \cos \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )-\sin \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )\right ) x}{c_{3} \sin \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )+\cos \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (c_{3} \cos \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )-\sin \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )\right ) x}{c_{3} \sin \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )+\cos \left (\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )+3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )+2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -y^{2} x^{2}-x^{4}+y^{\prime } \ln \left (x -1\right ) x -y^{2} x -x^{3}-y \ln \left (x -1\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y \ln \left (x -1\right )+x^{4}+x^{3}+y^{2} x^{2}+y^{2} x}{\ln \left (x -1\right ) x} \end {array} \]
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 homogeneous D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 39
dsolve(diff(y(x),x) = (y(x)*ln(x-1)+x^4+x^3+x^2*y(x)^2+x*y(x)^2)/ln(x-1)/x,y(x), singsol=all)
\[ y \left (x \right ) = \tan \left (-\operatorname {expIntegral}_{1}\left (-3 \ln \left (x -1\right )\right )-3 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (x -1\right )\right )-2 \,\operatorname {expIntegral}_{1}\left (-\ln \left (x -1\right )\right )+c_{1} \right ) x \]
✓ Solution by Mathematica
Time used: 0.86 (sec). Leaf size: 34
DSolve[y'[x] == (x^3 + x^4 + Log[-1 + x]*y[x] + x*y[x]^2 + x^2*y[x]^2)/(x*Log[-1 + x]),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x \tan (2 \operatorname {ExpIntegralEi}(\log (x-1))+3 \operatorname {ExpIntegralEi}(2 \log (x-1))+\operatorname {ExpIntegralEi}(3 \log (x-1))+c_1) \]