2.16   ODE No. 16

1. Problem in Latex
2. Mathematica input
3. Maple input

$$f(x)(xy(x)-1)+y^{\prime }(x)+y(x{)}^2=0$$ ✓ Mathematica : cpu = 0.0372842 (sec), leaf count = 186

$$\left\{\{y(x)\to \frac{\exp ({\int }_1^x\frac{f(K[1])K[1{]}^2+2}{K[1]}dK[1]){\int }_1^x\exp (-{\int }_1^{K[2]}\frac{f(K[1])K[1{]}^2+2}{K[1]}dK[1])dK[2]+c_1\exp ({\int }_1^x\frac{f(K[1])K[1{]}^2+2}{K[1]}dK[1])+x}{x(\exp ({\int }_1^x\frac{f(K[1])K[1{]}^2+2}{K[1]}dK[1]){\int }_1^x\exp (-{\int }_1^{K[2]}\frac{f(K[1])K[1{]}^2+2}{K[1]}dK[1])dK[2]+c_1\exp ({\int }_1^x\frac{f(K[1])K[1{]}^2+2}{K[1]}dK[1]))}\}\right\}$$ ✓ Maple : cpu = 0.088 (sec), leaf count = 49

$$\{y\left(x\right)={\mathrm{e}}^{\int \frac{-x^{2}f\left(x\right)-2}{x}\mathrm{d}x}{(-\mathit{\_}\mathit{C}\mathit{1}+\int {\mathrm{e}}^{\int \frac{-x^{2}f\left(x\right)-2}{x}\mathrm{d}x}\mathrm{d}x)}^{-1}+x^{-1}\}$$

Hand solution

$$\begin{array}{rl}{y}^{\prime }+y^2+(xy-1)f& =0\\ \text{(1)}& y^{\prime }\left(x\right)& =(-xy+1)f-y^2\end{array}$$

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$$\begin{array}{rl}y\left(x\right)& =y_p\left(x\right)+\frac{1}{u\left(x\right)}\\ & =\frac{1}{x}+\frac{1}{u}\end{array}$$

Hence$$\begin{array}{rl}{y}^{\prime }\left(x\right)& =y_p^{\prime }\left(x\right)-\frac{u^{\prime }\left(x\right)}{u^2\left(x\right)}\\ \text{(2)}& & =\frac{-1}{x^2}-\frac{u^{\prime }\left(x\right)}{u^2\left(x\right)}\end{array}$$

Equating (1) and (2) gives$$\begin{array}{rl}(-xy+1)f-y^2& =\frac{-1}{x^2}-\frac{u^{\prime }}{u^2}\\ (-x(\frac{1}{x}+\frac{1}{u})+1)f-{(\frac{1}{x}+\frac{1}{u})}^2& =\frac{-1}{x^2}-\frac{u^{\prime }}{u^2}\\ ((-1-\frac{x}{u})+1)f-(\frac{1}{x^2}+\frac{1}{u^2}+\frac{2}{xu})& =\frac{-1}{x^2}-\frac{u^{\prime }}{u^2}\\ -\frac{x}{u}f-(\frac{1}{x^2}+\frac{1}{u^2}+\frac{2}{xu})& =\frac{-1}{x^2}-\frac{u^{\prime }}{u^2}\\ -\frac{x}{u}f-\frac{1}{x^2}-\frac{1}{u^2}-\frac{2}{xu}& =\frac{-1}{x^2}-\frac{u^{\prime }}{u^2}\\ -xuf-1-\frac{2u}{x}& =-u^{\prime }\\ u^{\prime }& =xuf+1+\frac{2u}{x}\end{array}$$

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

Hence$$\begin{array}{rl}y& =y_p+\frac{1}{u}\\ & =\frac{1}{x}+\frac{1}{e^{-\int (xf+\frac{2}{x})dx}(\int e^{\int (xf+\frac{2}{x})dx}dx+C)}\end{array}$$

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