2.16   ODE No. 16

\[ f(x) (x y(x)-1)+y'(x)+y(x)^2=0 \] Mathematica : cpu = 0.0789155 (sec), leaf count = 186


\[\left \{\left \{y(x)\to \frac {\exp \left (\int _1^x\frac {f(K[1]) K[1]^2+2}{K[1]}dK[1]\right ) \int _1^x\exp \left (-\int _1^{K[2]}\frac {f(K[1]) K[1]^2+2}{K[1]}dK[1]\right )dK[2]+c_1 \exp \left (\int _1^x\frac {f(K[1]) K[1]^2+2}{K[1]}dK[1]\right )+x}{x \left (\exp \left (\int _1^x\frac {f(K[1]) K[1]^2+2}{K[1]}dK[1]\right ) \int _1^x\exp \left (-\int _1^{K[2]}\frac {f(K[1]) K[1]^2+2}{K[1]}dK[1]\right )dK[2]+c_1 \exp \left (\int _1^x\frac {f(K[1]) K[1]^2+2}{K[1]}dK[1]\right )\right )}\right \}\right \}\] Maple : cpu = 0.104 (sec), leaf count = 49


\[y \relax (x ) = \frac {{\mathrm e}^{\int \frac {-x^{2} f \relax (x )-2}{x}d x}}{-c_{1}+\int {\mathrm e}^{\int \frac {-x^{2} f \relax (x )-2}{x}d x}d x}+\frac {1}{x}\]

Hand solution

\begin {align} y^{\prime }+y^{2}+\left (xy-1\right ) f & =0\nonumber \\ y^{\prime }\relax (x) & =\left (-xy+1\right ) f-y^{2}\tag {1} \end {align}

This is Riccati first order non-linear ODE of the form. We can see a particular solution is \(y_{p}=\frac {1}{x}\), therefore, we use the substitution\begin {align*} y\relax (x) & =y_{p}\relax (x) +\frac {1}{u\relax (x) }\\ & =\frac {1}{x}+\frac {1}{u} \end {align*}

Hence\begin {align} y^{\prime }\relax (x) & =y_{p}^{\prime }\relax (x) -\frac {u^{\prime }\relax (x) }{u^{2}\relax (x) }\nonumber \\ & =\frac {-1}{x^{2}}-\frac {u^{\prime }\relax (x) }{u^{2}\relax (x) }\tag {2} \end {align}

Equating (1) and (2) gives\begin {align*} \left (-xy+1\right ) f-y^{2} & =\frac {-1}{x^{2}}-\frac {u^{\prime }}{u^{2}}\\ \left (-x\left (\frac {1}{x}+\frac {1}{u}\right ) +1\right ) f-\left ( \frac {1}{x}+\frac {1}{u}\right ) ^{2} & =\frac {-1}{x^{2}}-\frac {u^{\prime }}{u^{2}}\\ \left (\left (-1-\frac {x}{u}\right ) +1\right ) f-\left (\frac {1}{x^{2}}+\frac {1}{u^{2}}+\frac {2}{xu}\right ) & =\frac {-1}{x^{2}}-\frac {u^{\prime }}{u^{2}}\\ -\frac {x}{u}f-\left (\frac {1}{x^{2}}+\frac {1}{u^{2}}+\frac {2}{xu}\right ) & =\frac {-1}{x^{2}}-\frac {u^{\prime }}{u^{2}}\\ -\frac {x}{u}f-\frac {1}{x^{2}}-\frac {1}{u^{2}}-\frac {2}{xu} & =\frac {-1}{x^{2}}-\frac {u^{\prime }}{u^{2}}\\ -xuf-1-\frac {2u}{x} & =-u^{\prime }\\ u^{\prime } & =xuf+1+\frac {2u}{x} \end {align*}

Hence\[ u^{\prime }-\left (xf+\frac {2}{x}\right ) u=1 \] Integrating factor is \(\mu =e^{\int \left (xf+\frac {2}{x}\right ) dx}\), hence the solution is\[ d\left (\mu u\right ) =\mu \] Integrating both sides\begin {align*} \mu u & =\int \mu dx+C\\ u & =e^{-\int \left (xf+\frac {2}{x}\right ) dx}\int e^{\int \left ( xf+\frac {2}{x}\right ) dx}dx+Ce^{-\int \left (xf+\frac {2}{x}\right ) dx}\\ & =e^{-\int \left (xf+\frac {2}{x}\right ) dx}\left (\int e^{\int \left ( xf+\frac {2}{x}\right ) dx}dx+C\right ) \end {align*}

Hence\begin {align*} y & =y_{p}+\frac {1}{u}\\ & =\frac {1}{x}+\frac {1}{e^{-\int \left (xf+\frac {2}{x}\right ) dx}\left (\int e^{\int \left (xf+\frac {2}{x}\right ) dx}dx+C\right ) } \end {align*}

Hence\[ y\relax (x) =\frac {1}{x}+e^{\int \left (xf+\frac {2}{x}\right ) dx}\left (\int e^{\int \left (xf+\frac {2}{x}\right ) dx}dx+C\right ) ^{-1}\]