Internal problem ID [9030]
Internal file name [OUTPUT/7965_Monday_June_06_2022_01_02_21_AM_34294899/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 696.
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]
\[ \boxed {y^{\prime }-\frac {y \ln \left (x -1\right )+{\mathrm e}^{x +1} x^{3}+7 \,{\mathrm e}^{x +1} x y^{2}}{\ln \left (x -1\right ) x}=0} \]
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*} 7 \,{\mathrm e}^{x +1} x^{3} u \left (x \right )^{2}+{\mathrm e}^{x +1} x^{3}-\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) \ln \left (x -1\right ) x +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 {{\mathrm e}^{x +1} x \left (7 u^{2}+1\right )}{\ln \left (x -1\right )} \end {align*}
Where \(f(x)=\frac {{\mathrm e}^{x +1} x}{\ln \left (x -1\right )}\) and \(g(u)=7 u^{2}+1\). Integrating both sides gives \begin{align*} \frac {1}{7 u^{2}+1} \,du &= \frac {{\mathrm e}^{x +1} x}{\ln \left (x -1\right )} \,d x \\ \int { \frac {1}{7 u^{2}+1} \,du} &= \int {\frac {{\mathrm e}^{x +1} x}{\ln \left (x -1\right )} \,d x} \\ \frac {\sqrt {7}\, \arctan \left (u \sqrt {7}\right )}{7}&=\int \frac {{\mathrm e}^{x +1} x}{\ln \left (x -1\right )}d x +c_{2} \\ \end{align*} The solution is \[ \frac {\sqrt {7}\, \arctan \left (u \left (x \right ) \sqrt {7}\right )}{7}-\left (\int \frac {{\mathrm e}^{x +1} x}{\ln \left (x -1\right )}d x \right )-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {\sqrt {7}\, \arctan \left (\frac {y \sqrt {7}}{x}\right )}{7}-\left (\int \frac {{\mathrm e}^{x +1} x}{\ln \left (x -1\right )}d x \right )-c_{2} = 0\\ \frac {\sqrt {7}\, \arctan \left (\frac {y \sqrt {7}}{x}\right )}{7}-\left (\int \frac {{\mathrm e}^{x +1} x}{\ln \left (x -1\right )}d x \right )-c_{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {\sqrt {7}\, \arctan \left (\frac {y \sqrt {7}}{x}\right )}{7}-\left (\int \frac {{\mathrm e}^{x +1} x}{\ln \left (x -1\right )}d x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {\sqrt {7}\, \arctan \left (\frac {y \sqrt {7}}{x}\right )}{7}-\left (\int \frac {{\mathrm e}^{x +1} x}{\ln \left (x -1\right )}d x \right )-c_{2} = 0 \] Verified OK.
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y \ln \left (x -1\right )+{\mathrm e}^{x +1} x^{3}+7 \,{\mathrm e}^{x +1} x \,y^{2}}{\ln \left (x -1\right ) x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {y}{x}+\frac {x^{2} {\mathrm e}^{x} {\mathrm e}}{\ln \left (x -1\right )}+\frac {7 \,{\mathrm e}^{x} {\mathrm e} y^{2}}{\ln \left (x -1\right )} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {x^{2} {\mathrm e}^{x +1}}{\ln \left (x -1\right )}\), \(f_1(x)=\frac {1}{x}\) and \(f_2(x)=\frac {7 \,{\mathrm e}^{x +1}}{\ln \left (x -1\right )}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {7 \,{\mathrm e}^{x +1} u}{\ln \left (x -1\right )}} \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 {7 \,{\mathrm e}^{x +1}}{\ln \left (x -1\right )}-\frac {7 \,{\mathrm e}^{x +1}}{\ln \left (x -1\right )^{2} \left (x -1\right )}\\ f_1 f_2 &=\frac {7 \,{\mathrm e}^{x +1}}{x \ln \left (x -1\right )}\\ f_2^2 f_0 &=\frac {49 \,{\mathrm e}^{3 x +3} x^{2}}{\ln \left (x -1\right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {7 \,{\mathrm e}^{x +1} u^{\prime \prime }\left (x \right )}{\ln \left (x -1\right )}-\left (\frac {7 \,{\mathrm e}^{x +1}}{\ln \left (x -1\right )}-\frac {7 \,{\mathrm e}^{x +1}}{\ln \left (x -1\right )^{2} \left (x -1\right )}+\frac {7 \,{\mathrm e}^{x +1}}{x \ln \left (x -1\right )}\right ) u^{\prime }\left (x \right )+\frac {49 \,{\mathrm e}^{3 x +3} x^{2} u \left (x \right )}{\ln \left (x -1\right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{-\sqrt {7}\, \left (\int \frac {\tan \left (\sqrt {7}\, {\mathrm e} \left (\int \frac {x \,{\mathrm e}^{x}}{\ln \left (x -1\right )}d x \right )+c_{1} \right ) {\mathrm e}^{x +1} x}{\ln \left (x -1\right )}d x \right )} c_{2} \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {x c_{2} \sqrt {7}\, {\mathrm e}^{x +1-\sqrt {7}\, \left (\int \frac {\tan \left (\sqrt {7}\, {\mathrm e} \left (\int \frac {x \,{\mathrm e}^{x}}{\ln \left (x -1\right )}d x \right )+c_{1} \right ) {\mathrm e}^{x +1} x}{\ln \left (x -1\right )}d x \right )} \tan \left (\sqrt {7}\, {\mathrm e} \left (\int \frac {x \,{\mathrm e}^{x}}{\ln \left (x -1\right )}d x \right )+c_{1} \right )}{\ln \left (x -1\right )} \] Using the above in (1) gives the solution \[ y = \frac {x \sqrt {7}\, {\mathrm e}^{x +1-\sqrt {7}\, \left (\int \frac {\tan \left (\sqrt {7}\, {\mathrm e} \left (\int \frac {x \,{\mathrm e}^{x}}{\ln \left (x -1\right )}d x \right )+c_{1} \right ) {\mathrm e}^{x +1} x}{\ln \left (x -1\right )}d x \right )} \tan \left (\sqrt {7}\, {\mathrm e} \left (\int \frac {x \,{\mathrm e}^{x}}{\ln \left (x -1\right )}d x \right )+c_{1} \right ) {\mathrm e}^{-x -1} {\mathrm e}^{\int \frac {\tan \left (\int \frac {\sqrt {7}\, x \,{\mathrm e}^{x +1}}{\ln \left (x -1\right )}d x +c_{1} \right ) {\mathrm e}^{x +1} \sqrt {7}\, x}{\ln \left (x -1\right )}d x}}{7} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {x \sqrt {7}\, \tan \left (\sqrt {7}\, {\mathrm e} \left (\int \frac {x \,{\mathrm e}^{x}}{\ln \left (x -1\right )}d x \right )+c_{3} \right )}{7} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x \sqrt {7}\, \tan \left (\sqrt {7}\, {\mathrm e} \left (\int \frac {x \,{\mathrm e}^{x}}{\ln \left (x -1\right )}d x \right )+c_{3} \right )}{7} \\ \end{align*}
Verification of solutions
\[ y = \frac {x \sqrt {7}\, \tan \left (\sqrt {7}\, {\mathrm e} \left (\int \frac {x \,{\mathrm e}^{x}}{\ln \left (x -1\right )}d x \right )+c_{3} \right )}{7} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 7 \,{\mathrm e}^{x +1} x y^{2}+{\mathrm e}^{x +1} x^{3}-y^{\prime } \ln \left (x -1\right ) x +y \ln \left (x -1\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {-y \ln \left (x -1\right )-{\mathrm e}^{x +1} x^{3}-7 \,{\mathrm e}^{x +1} x y^{2}}{\ln \left (x -1\right ) x} \end {array} \]
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: 0.047 (sec). Leaf size: 32
dsolve(diff(y(x),x) = (y(x)*ln(x-1)+exp(x+1)*x^3+7*exp(x+1)*x*y(x)^2)/ln(x-1)/x,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\tan \left (\left ({\mathrm e} \left (\int \frac {x \,{\mathrm e}^{x}}{\ln \left (x -1\right )}d x \right )+c_{1} \right ) \sqrt {7}\right ) x \sqrt {7}}{7} \]
✓ Solution by Mathematica
Time used: 1.53 (sec). Leaf size: 45
DSolve[y'[x] == (E^(1 + x)*x^3 + Log[-1 + x]*y[x] + 7*E^(1 + x)*x*y[x]^2)/(x*Log[-1 + x]),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {x \tan \left (\sqrt {7} \left (\int _1^x\frac {e^{K[1]+1} K[1]}{\log (K[1]-1)}dK[1]+c_1\right )\right )}{\sqrt {7}} \]