Internal problem ID [9038]
Internal file name [OUTPUT/7973_Monday_June_06_2022_01_03_50_AM_69612722/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 704.
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]
\[ \boxed {y^{\prime }-\frac {y \ln \left (x \right ) x -y+2 b \,x^{5}+2 x^{3} a y^{2}}{\left (x \ln \left (x \right )-1\right ) x}=0} \]
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*} 2 x^{5} a u \left (x \right )^{2}+2 b \,x^{5}-\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) \ln \left (x \right ) x^{2}+u \left (x \right ) x^{2} \ln \left (x \right )+\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x -u \left (x \right ) x = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {x^{3} \left (2 a \,u^{2}+2 b \right )}{x \ln \left (x \right )-1} \end {align*}
Where \(f(x)=\frac {x^{3}}{x \ln \left (x \right )-1}\) and \(g(u)=2 a \,u^{2}+2 b\). Integrating both sides gives \begin{align*} \frac {1}{2 a \,u^{2}+2 b} \,du &= \frac {x^{3}}{x \ln \left (x \right )-1} \,d x \\ \int { \frac {1}{2 a \,u^{2}+2 b} \,du} &= \int {\frac {x^{3}}{x \ln \left (x \right )-1} \,d x} \\ \frac {\arctan \left (\frac {a u}{\sqrt {a b}}\right )}{2 \sqrt {a b}}&=\int \frac {x^{3}}{x \ln \left (x \right )-1}d x +c_{2} \\ \end{align*} The solution is \[ \frac {\arctan \left (\frac {a u \left (x \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}-\left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {\arctan \left (\frac {y a}{x \sqrt {a b}}\right )}{2 \sqrt {a b}}-\left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )-c_{2} = 0\\ \frac {\arctan \left (\frac {y a}{x \sqrt {a b}}\right )}{2 \sqrt {a b}}-\left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )-c_{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {\arctan \left (\frac {y a}{x \sqrt {a b}}\right )}{2 \sqrt {a b}}-\left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {\arctan \left (\frac {y a}{x \sqrt {a b}}\right )}{2 \sqrt {a b}}-\left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )-c_{2} = 0 \] Verified OK.
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y x \ln \left (x \right )-y +2 b \,x^{5}+2 x^{3} a \,y^{2}}{\left (x \ln \left (x \right )-1\right ) x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {2 x^{2} a \,y^{2}}{x \ln \left (x \right )-1}+\frac {2 x^{4} b}{x \ln \left (x \right )-1}+\frac {y \ln \left (x \right )}{x \ln \left (x \right )-1}-\frac {y}{\left (x \ln \left (x \right )-1\right ) x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {2 x^{4} b}{x \ln \left (x \right )-1}\), \(f_1(x)=\frac {1}{x}\) and \(f_2(x)=\frac {2 a \,x^{2}}{x \ln \left (x \right )-1}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {2 a \,x^{2} u}{x \ln \left (x \right )-1}} \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 {4 a x}{x \ln \left (x \right )-1}-\frac {2 a \,x^{2} \left (1+\ln \left (x \right )\right )}{\left (x \ln \left (x \right )-1\right )^{2}}\\ f_1 f_2 &=\frac {2 a x}{x \ln \left (x \right )-1}\\ f_2^2 f_0 &=\frac {8 a^{2} x^{8} b}{\left (x \ln \left (x \right )-1\right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {2 a \,x^{2} u^{\prime \prime }\left (x \right )}{x \ln \left (x \right )-1}-\left (\frac {6 a x}{x \ln \left (x \right )-1}-\frac {2 a \,x^{2} \left (1+\ln \left (x \right )\right )}{\left (x \ln \left (x \right )-1\right )^{2}}\right ) u^{\prime }\left (x \right )+\frac {8 a^{2} x^{8} b u \left (x \right )}{\left (x \ln \left (x \right )-1\right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} {\mathrm e}^{2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}+c_{2} {\mathrm e}^{-2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {2 i x^{3} \sqrt {a b}\, \left (c_{1} {\mathrm e}^{2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}-c_{2} {\mathrm e}^{-2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}\right )}{x \ln \left (x \right )-1} \] Using the above in (1) gives the solution \[ y = -\frac {i x \sqrt {a b}\, \left (c_{1} {\mathrm e}^{2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}-c_{2} {\mathrm e}^{-2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}\right )}{a \left (c_{1} {\mathrm e}^{2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}+c_{2} {\mathrm e}^{-2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {i x \sqrt {a b}\, \left (c_{1} {\mathrm e}^{2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}-c_{2} {\mathrm e}^{-2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}\right )}{a \left (c_{1} {\mathrm e}^{2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}+c_{2} {\mathrm e}^{-2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {i x \sqrt {a b}\, \left (c_{1} {\mathrm e}^{2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}-c_{2} {\mathrm e}^{-2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}\right )}{a \left (c_{1} {\mathrm e}^{2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}+c_{2} {\mathrm e}^{-2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {i x \sqrt {a b}\, \left (c_{1} {\mathrm e}^{2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}-c_{2} {\mathrm e}^{-2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}\right )}{a \left (c_{1} {\mathrm e}^{2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}+c_{2} {\mathrm e}^{-2 i \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x \right )}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x^{3} a y^{2}+2 b \,x^{5}-y^{\prime } \ln \left (x \right ) x^{2}+y \ln \left (x \right ) x +y^{\prime } x -y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-y \ln \left (x \right ) x +y-2 b \,x^{5}-2 x^{3} a y^{2}}{-x^{2} \ln \left (x \right )+x} \end {array} \]
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: 38
dsolve(diff(y(x),x) = (y(x)*ln(x)*x-y(x)+2*x^5*b+2*x^3*a*y(x)^2)/(x*ln(x)-1)/x,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\tan \left (2 \sqrt {a b}\, \left (\int \frac {x^{3}}{x \ln \left (x \right )-1}d x +c_{1} \right )\right ) x \sqrt {a b}}{a} \]
✓ Solution by Mathematica
Time used: 53.989 (sec). Leaf size: 55
DSolve[y'[x] == (2*b*x^5 - y[x] + x*Log[x]*y[x] + 2*a*x^3*y[x]^2)/(x*(-1 + x*Log[x])),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {\sqrt {b} x \tan \left (\sqrt {a} \sqrt {b} \left (\int _1^x\frac {2 K[1]^3}{K[1] \log (K[1])-1}dK[1]+c_1\right )\right )}{\sqrt {a}} \]