2.20   ODE No. 20

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

$$(x^2+1)y(x)+y^{\prime }(x)-y(x{)}^2-2x=0$$ ✓ Mathematica : cpu = 0.0774186 (sec), leaf count = 49

$$\left\{\{y(x)\to \frac{e^{\frac{x^3}{3}+x}}{-{\int }_1^x{e}^{\frac{K[1{]}^3}{3}+K[1]}dK[1]+c_1}+x^2+1\}\right\}$$ ✓ Maple : cpu = 0.054 (sec), leaf count = 34

$$\{y\left(x\right)=x^2+1+{\mathrm{e}}^{\frac{x^3}{3}+x}{(\mathit{\_}\mathit{C}\mathit{1}-\int {\mathrm{e}}^{\frac{x^3}{3}+x}\mathrm{d}x)}^{-1}\}$$

Hand solution

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

This is Riccati first order non-linear ODE of the form of the general form $y^{\prime }=P\left(x\right)+Q\left(x\right)y+R\left(x\right)y^2$ where $P\left(x\right)=2x,Q\left(x\right)=-(x^2+1),R\left(x\right)=1$. We can convert this to Bernoulli first order ODE in $u\left(x\right)$, which is little easier to solve by using $u=y-x^2-1$.  The difference between Bernoulli and Riccati is that the term $P\left(x\right)=0$ in Bernoulli. If $P\left(x\right)\ne 0$ and $R\left(x\right)\ne 0$ then it is called Riccati.

Using $u=y-x^2-1$ gives$$\begin{array}{rl}{u}^{\prime }& =y^{\prime }-2x\\ u^{\prime }& =[-(x^2+1)y+y^2+2x]-2x\\ & =-(x^2+1)(u+x^2+1)+{(u+x^2+1)}^2\\ & =(u+x^2+1)[(u+x^2+1)-(x^2+1)]\\ & =(u+x^2+1)u\\ & =u^2+u(1+x^2)\end{array}$$

We see now this is Bernoulli since $P\left(x\right)=0$. To solve Bernoulli we always start by dividing by $u^2$ giving$$\frac{u^{\prime }}{u^2}=1+\frac{1}{u}(1+x^2)$$ Next we let $v=\frac{1}{u}$, hence $v^{\prime }=-\frac{u^{\prime }}{u^2}$therefore the above becomes$$\begin{array}{rl}-v^{\prime }& =1+v(1+x^2)\\ v^{\prime }+v(1+x^2)& =-1\end{array}$$

Integrating factor is $e^{\int (1+x^2)dx}=e^{(x+\frac{x^3}{2})}$, therefore$$d\left(e^{(x+\frac{x^3}{2})}v\right)=-e^{(x+\frac{x^3}{2})}$$ Integrating$$\begin{array}{rl}{e}^{(x+\frac{x^3}{2})}v& =-\int e^{(x+\frac{x^3}{2})}dx+C\\ v\left(x\right)& =e^{-(x+\frac{x^3}{2})}(C-\int e^{(x+\frac{x^3}{2})}dx)\end{array}$$

Therefore $$u=\frac{1}{v}=\frac{e^{(x+\frac{x^3}{2})}}{(C-\int e^{(x+\frac{x^3}{2})}dx)}$$ And since $u=y-x^2-1$ then$$\begin{array}{rl}y\left(x\right)& =u+1+x^2\\ & =\frac{e^{(x+\frac{x^3}{2})}}{(C-\int e^{(x+\frac{x^3}{2})}dx)}+1+x^2\end{array}$$