Internal problem ID [9015]
Internal file name [OUTPUT/7950_Monday_June_06_2022_01_00_08_AM_55467881/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 681.
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]
\[ \boxed {y^{\prime }-\frac {y+x^{3} b \ln \left (\frac {1}{x}\right )+x^{4} b +b \,x^{3}+x a y^{2} \ln \left (\frac {1}{x}\right )+y^{2} a \,x^{2}+y^{2} a x}{x}=0} \]
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*} x^{3} a u \left (x \right )^{2} \ln \left (\frac {1}{x}\right )+x^{3} b \ln \left (\frac {1}{x}\right )+u \left (x \right )^{2} x^{4} a +x^{4} b +u \left (x \right )^{2} x^{3} a +b \,x^{3}-\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)\\ &= x \left (u^{2} a +b \right ) \left (\ln \left (\frac {1}{x}\right )+x +1\right ) \end {align*}
Where \(f(x)=x \left (\ln \left (\frac {1}{x}\right )+x +1\right )\) and \(g(u)=u^{2} a +b\). Integrating both sides gives \begin{align*} \frac {1}{u^{2} a +b} \,du &= x \left (\ln \left (\frac {1}{x}\right )+x +1\right ) \,d x \\ \int { \frac {1}{u^{2} a +b} \,du} &= \int {x \left (\ln \left (\frac {1}{x}\right )+x +1\right ) \,d x} \\ \frac {\arctan \left (\frac {u a}{\sqrt {a b}}\right )}{\sqrt {a b}}&=\frac {x^{2} \ln \left (\frac {1}{x}\right )}{2}+\frac {3 x^{2}}{4}+\frac {x^{3}}{3}+c_{2} \\ \end{align*} The solution is \[ \frac {\arctan \left (\frac {u \left (x \right ) a}{\sqrt {a b}}\right )}{\sqrt {a b}}-\frac {x^{2} \ln \left (\frac {1}{x}\right )}{2}-\frac {3 x^{2}}{4}-\frac {x^{3}}{3}-c_{2} = 0 \] 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 )}{\sqrt {a b}}-\frac {x^{2} \ln \left (\frac {1}{x}\right )}{2}-\frac {3 x^{2}}{4}-\frac {x^{3}}{3}-c_{2} = 0\\ \frac {\arctan \left (\frac {y a}{x \sqrt {a b}}\right )}{\sqrt {a b}}-\frac {x^{2} \ln \left (\frac {1}{x}\right )}{2}-\frac {3 x^{2}}{4}-\frac {x^{3}}{3}-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 )}{\sqrt {a b}}-\frac {x^{2} \ln \left (\frac {1}{x}\right )}{2}-\frac {3 x^{2}}{4}-\frac {x^{3}}{3}-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {\arctan \left (\frac {y a}{x \sqrt {a b}}\right )}{\sqrt {a b}}-\frac {x^{2} \ln \left (\frac {1}{x}\right )}{2}-\frac {3 x^{2}}{4}-\frac {x^{3}}{3}-c_{2} = 0 \] Verified OK.
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y +x^{3} b \ln \left (\frac {1}{x}\right )+x^{4} b +b \,x^{3}+x a \,y^{2} \ln \left (\frac {1}{x}\right )+a \,x^{2} y^{2}+a x \,y^{2}}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,y^{2} \ln \left (\frac {1}{x}\right )+x^{2} b \ln \left (\frac {1}{x}\right )+a x \,y^{2}+b \,x^{3}+y^{2} a +b \,x^{2}+\frac {y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {x^{3} b \ln \left (\frac {1}{x}\right )+x^{4} b +b \,x^{3}}{x}\), \(f_1(x)=\frac {1}{x}\) and \(f_2(x)=\frac {\ln \left (\frac {1}{x}\right ) a x +a \,x^{2}+a x}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {\left (\ln \left (\frac {1}{x}\right ) a x +a \,x^{2}+a x \right ) 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 {\ln \left (\frac {1}{x}\right ) a +2 a x}{x}-\frac {\ln \left (\frac {1}{x}\right ) a x +a \,x^{2}+a x}{x^{2}}\\ f_1 f_2 &=\frac {\ln \left (\frac {1}{x}\right ) a x +a \,x^{2}+a x}{x^{2}}\\ f_2^2 f_0 &=\frac {\left (\ln \left (\frac {1}{x}\right ) a x +a \,x^{2}+a x \right )^{2} \left (x^{3} b \ln \left (\frac {1}{x}\right )+x^{4} b +b \,x^{3}\right )}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {\left (\ln \left (\frac {1}{x}\right ) a x +a \,x^{2}+a x \right ) u^{\prime \prime }\left (x \right )}{x}-\frac {\left (\ln \left (\frac {1}{x}\right ) a +2 a x \right ) u^{\prime }\left (x \right )}{x}+\frac {\left (\ln \left (\frac {1}{x}\right ) a x +a \,x^{2}+a x \right )^{2} \left (x^{3} b \ln \left (\frac {1}{x}\right )+x^{4} b +b \,x^{3}\right ) u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \sin \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )+c_{2} \cos \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (c_{1} \cos \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )-c_{2} \sin \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )\right ) a b \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2}}{\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}} \] Using the above in (1) gives the solution \[ y = -\frac {\left (c_{1} \cos \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )-c_{2} \sin \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )\right ) a b \,x^{3} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2}}{\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, \left (\ln \left (\frac {1}{x}\right ) a x +a \,x^{2}+a x \right ) \left (c_{1} \sin \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )+c_{2} \cos \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (\ln \left (\frac {1}{x}\right )+x +1\right ) x^{2} b \left (-c_{3} \cos \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )+\sin \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )\right )}{\left (c_{3} \sin \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )+\cos \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )\right ) \sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\ln \left (\frac {1}{x}\right )+x +1\right ) x^{2} b \left (-c_{3} \cos \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )+\sin \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )\right )}{\left (c_{3} \sin \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )+\cos \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )\right ) \sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (\ln \left (\frac {1}{x}\right )+x +1\right ) x^{2} b \left (-c_{3} \cos \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )+\sin \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )\right )}{\left (c_{3} \sin \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )+\cos \left (\frac {\sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}\, x \left (6 \ln \left (\frac {1}{x}\right )+9+4 x \right )}{12 \ln \left (\frac {1}{x}\right )+12 x +12}\right )\right ) \sqrt {a \,x^{2} \left (\ln \left (\frac {1}{x}\right )+x +1\right )^{2} b}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x a y^{2} \ln \left (\frac {1}{x}\right )+x^{3} b \ln \left (\frac {1}{x}\right )+y^{2} a \,x^{2}+x^{4} b +y^{2} a x +b \,x^{3}-y^{\prime } x +y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y+x^{3} b \ln \left (\frac {1}{x}\right )+x^{4} b +b \,x^{3}+x a y^{2} \ln \left (\frac {1}{x}\right )+y^{2} a \,x^{2}+y^{2} a x}{x} \end {array} \]
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: 0.016 (sec). Leaf size: 45
dsolve(diff(y(x),x) = (y(x)+x^3*b*ln(1/x)+x^4*b+b*x^3+x*a*y(x)^2*ln(1/x)+x^2*a*y(x)^2+a*x*y(x)^2)/x,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\tan \left (\frac {\left (6 x^{2} \ln \left (\frac {1}{x}\right )+4 x^{3}+9 x^{2}+12 c_{1} \right ) \sqrt {a b}}{12}\right ) x \sqrt {a b}}{a} \]
✓ Solution by Mathematica
Time used: 43.49 (sec). Leaf size: 54
DSolve[y'[x] == (b*x^3 + b*x^4 + b*x^3*Log[x^(-1)] + y[x] + a*x*y[x]^2 + a*x^2*y[x]^2 + a*x*Log[x^(-1)]*y[x]^2)/x,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {\sqrt {b} x \tan \left (\frac {1}{12} \sqrt {a} \sqrt {b} \left (4 x^3+9 x^2-6 x^2 \log (x)+12 c_1\right )\right )}{\sqrt {a}} \]