Internal problem ID [8487]
Internal file name [OUTPUT/7420_Sunday_June_05_2022_10_54_49_PM_17516898/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 151.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "abelFirstKind"
Maple gives the following as the ode type
[_rational, _Abel]
Unable to solve or complete the solution.
\[ \boxed {\left (x^{2}+1\right ) y^{\prime }+\left (y^{2}+1\right ) \left (2 x y-1\right )=0} \]
This is Abel first kind ODE, it has the form \[ y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \] Comparing the above to given ODE which is \begin {align*} y^{\prime }&=-\frac {2 x y^{3}}{x^{2}+1}+\frac {y^{2}}{x^{2}+1}-\frac {2 x y}{x^{2}+1}+\frac {1}{x^{2}+1}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= \frac {1}{x^{2}+1}\\ f_1(x) &= -\frac {2 x}{x^{2}+1}\\ f_2(x) &= \frac {1}{x^{2}+1}\\ f_3(x) &= -\frac {2 x}{x^{2}+1} \end {align*}
Since \(f_2(x)=\frac {1}{x^{2}+1}\) is not zero, then the first step is to apply the following transformation to remove \(f_2\). Let \(y = u(x) - \frac {f_2}{3 f_3}\) or \begin {align*} y &= u(x) - \left ( \frac {\frac {1}{x^{2}+1}}{-\frac {6 x}{x^{2}+1}} \right ) \\ &= u \left (x \right )+\frac {1}{6 x} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = -\frac {2 x u \left (x \right )^{3}}{x^{2}+1}-\frac {2 x u \left (x \right )}{x^{2}+1}+\frac {u \left (x \right )}{6 \left (x^{2}+1\right ) x}+\frac {5}{6 \left (x^{2}+1\right )}+\frac {5}{27 \left (x^{2}+1\right ) x^{2}}\tag {2} \end {align*}
This is Abel first kind ODE, it has the form \[ u^{\prime }\left (x \right )= f_0(x)+f_1(x) u \left (x \right ) +f_2(x)u \left (x \right )^{2}+f_3(x)u \left (x \right )^{3} \] Comparing the above to given ODE which is \begin {align*} u^{\prime }\left (x \right )&=-\frac {2 x u \left (x \right )^{3}}{x^{2}+1}-\frac {\left (108 x^{3}-9 x \right ) u \left (x \right )}{54 \left (x^{2}+1\right ) x^{2}}-\frac {-45 x^{2}-10}{54 \left (x^{2}+1\right ) x^{2}}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= \frac {5}{6 \left (x^{2}+1\right )}+\frac {5}{27 \left (x^{2}+1\right ) x^{2}}\\ f_1(x) &= -\frac {2 x}{x^{2}+1}+\frac {1}{6 \left (x^{2}+1\right ) x}\\ f_2(x) &= 0\\ f_3(x) &= -\frac {2 x}{x^{2}+1} \end {align*}
Since \(f_2(x)=0\) then we check the Abel invariant to see if it depends on \(x\) or not. The Abel invariant is given by \begin {align*} -\frac {f_{1}^{3}}{f_{0}^{2} f_{3}} \end {align*}
Which when evaluating gives \begin {align*} -\frac {{\left (\frac {2 \left (-\frac {5 x}{3 \left (x^{2}+1\right )^{2}}-\frac {10}{27 \left (x^{2}+1\right )^{2} x}-\frac {10}{27 \left (x^{2}+1\right ) x^{3}}\right ) x}{x^{2}+1}+\left (\frac {5}{6 \left (x^{2}+1\right )}+\frac {5}{27 \left (x^{2}+1\right ) x^{2}}\right ) \left (\frac {4 x^{2}}{\left (x^{2}+1\right )^{2}}-\frac {2}{x^{2}+1}\right )-\frac {6 \left (\frac {5}{6 \left (x^{2}+1\right )}+\frac {5}{27 \left (x^{2}+1\right ) x^{2}}\right ) x \left (-\frac {2 x}{x^{2}+1}+\frac {1}{6 \left (x^{2}+1\right ) x}\right )}{x^{2}+1}\right )}^{3} \left (x^{2}+1\right )^{4}}{432 x^{4} {\left (\frac {5}{6 \left (x^{2}+1\right )}+\frac {5}{27 \left (x^{2}+1\right ) x^{2}}\right )}^{5}} \end {align*}
Since the Abel invariant depends on \(x\) then unable to solve this ode at this time.
Unable to complete the solution now.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) y^{\prime }+\left (y^{2}+1\right ) \left (2 x y-1\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\left (y^{2}+1\right ) \left (2 x y-1\right )}{x^{2}+1} \end {array} \]
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 differential order: 1; looking for linear symmetries trying exact trying Abel <- Abel successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 65
dsolve((x^2+1)*diff(y(x),x) + (y(x)^2+1)*(2*x*y(x) - 1)=0,y(x), singsol=all)
\[ c_{1} +\frac {x}{{\left (\frac {\left (x^{2}+1\right ) \left (y \left (x \right )^{2}+1\right )}{\left (x y \left (x \right )-1\right )^{2}}\right )}^{\frac {1}{4}}}+\frac {\left (x +y \left (x \right )\right ) \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {5}{4}\right ], \left [\frac {3}{2}\right ], -\frac {\left (x +y \left (x \right )\right )^{2}}{\left (x y \left (x \right )-1\right )^{2}}\right )}{2 x y \left (x \right )-2} = 0 \]
✓ Solution by Mathematica
Time used: 0.421 (sec). Leaf size: 203
DSolve[(x^2+1)*y'[x] + (y[x]^2+1)*(2*x*y[x] - 1)==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [c_1=\frac {\frac {1}{2} \left (\frac {1}{\frac {i x}{x^2+1}-\frac {i x^2 y(x)}{x^2+1}}+\frac {i}{x}\right ) \sqrt [4]{1-\left (\frac {1}{\frac {i x}{x^2+1}-\frac {i x^2 y(x)}{x^2+1}}+\frac {i}{x}\right )^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{4},\frac {3}{2},\left (\frac {1}{\frac {i x}{x^2+1}-\frac {i x^2 y(x)}{x^2+1}}+\frac {i}{x}\right )^2\right )+i x}{\sqrt [4]{-1+\left (\frac {1}{\frac {i x}{x^2+1}-\frac {i x^2 y(x)}{x^2+1}}+\frac {i}{x}\right )^2}},y(x)\right ] \]