2.27   ODE No. 27

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

$$ay(x)(y(x)-x)+y^{\prime }(x)-1=0$$ ✓ Mathematica : cpu = 2.64591 (sec), leaf count = 88

$$\left\{\{y(x)\to \frac{\sqrt{2\pi }{c}_{1}x\text{erf}\left(\frac{\sqrt{a}x}{\sqrt{2}}\right)+\frac{2(c_1{e}^{-\frac{a{x}^2}{2}}+ax)}{\sqrt{a}}}{\sqrt{2\pi }{c}_1\text{erf}\left(\frac{\sqrt{a}x}{\sqrt{2}}\right)+2\sqrt{a}}\}\right\}$$

✓ Maple : cpu = 0.418 (sec), leaf count = 72

$$\{y\left(x\right)=1(2\sqrt{a}{\mathrm{e}}^{-1/2a{x}^2}+x(\sqrt{\pi }\mathit{E}\mathit{r}\mathit{f}\left(\frac{\sqrt{2}x}{2}\sqrt{a}\right)\sqrt{2}a+2a^{3/2}\mathit{\_}\mathit{C}\mathit{1})){(\sqrt{\pi }\mathit{E}\mathit{r}\mathit{f}\left(\frac{\sqrt{2}x}{2}\sqrt{a}\right)\sqrt{2}a+2a^{3/2}\mathit{\_}\mathit{C}\mathit{1})}^{-1}\}$$

Hand solution

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

This is Riccati first order non-linear ODE $y^{\prime }=P\left(x\right)+A\left(x\right)y+R\left(x\right)y^2$ with $P\left(x\right)=1,Q\left(x\right)=-ax,R\left(x\right)=-a$. We can convert Riccati to Bernoulli which is easier to solve using the substitution $u=y-x$$$\begin{array}{rl}{u}^{\prime }& =y^{\prime }-1\\ & =(1+ayx-a{y}^2)-1\\ & =(1+a(u+x)x-a{(u+x)}^2)-1\\ & =1+aux+a{x}^2-a(u^2+x^2+2ux)-1\\ & =1+aux+a{x}^2-a{u}^2-a{x}^2-2aux-1\\ & =-aux-a{u}^2\\ u^{\prime }& =-aux-a{u}^2\end{array}$$

This is of the form $u^{\prime }=P\left(x\right)+Q\left(x\right)u+R\left(x\right)u^2$ and since $P\left(x\right)=0$ then it is Bernoulli differential equation. (when $P\left(x\right)\ne 0$ and $R\left(x\right)\ne 0$ it is Riccati). To solve Bernoulli we always start by dividing by $u^2$$$\frac{u^{\prime }}{u^2}=-\frac{ax}{u}-a$$ Then we let $\zeta =\frac{1}{u}$, hence ${\zeta }^{\prime }=-\frac{u^{\prime }}{u^2}$, therefore the above becomes$$\begin{array}{rl}-{\zeta }^{\prime }& =-ax\zeta -a\\ {\zeta }^{\prime }-ax\zeta & =a\end{array}$$

Integrating factor is $e^{-\int axdx}=e^{-a\frac{x^2}{2}}$, hence $d\left(e^{-a\frac{x^2}{2}}\zeta \right)=a{e}^{-a\frac{x^2}{2}}$. Integrating both sides gives$$e^{-a\frac{x^2}{2}}\zeta =a\int e^{-a\frac{x^2}{2}}dx+C$$ But $$\int e^{-a\frac{x^2}{2}}dx=\sqrt{\frac{\pi }{2a}}\mathrm{erf}\left(\sqrt{\frac{a}{2}}x\right)$$ Therefore$$\begin{array}{rl}{e}^{-a\frac{x^2}{2}}\zeta & =a\sqrt{\frac{\pi }{2a}}\mathrm{erf}\left(\sqrt{\frac{a}{2}}x\right)+C\\ \zeta & =e^{a\frac{x^2}{2}}(a\sqrt{\frac{\pi }{2a}}\mathrm{erf}\left(\sqrt{\frac{a}{2}}x\right)+C)\end{array}$$

Hence$$\begin{array}{rl}u& =\frac{1}{\zeta }\\ & =e^{-a\frac{x^2}{2}}{(a\sqrt{\frac{\pi }{2a}}\mathrm{erf}\left(\sqrt{\frac{a}{2}}x\right)+C)}^{-1}\end{array}$$

Since $u=y-x$ then$$\begin{array}{rl}y& =u+x\\ & =e^{-a\frac{x^2}{2}}{(a\sqrt{\frac{\pi }{2a}}\mathrm{erf}\left(\sqrt{\frac{a}{2}}x\right)+C)}^{-1}+x\\ & =\frac{e^{-a\frac{x^2}{2}}}{\sqrt{\frac{a\pi }{2}}\mathrm{erf}\left(\sqrt{\frac{a}{2}}x\right)+C}+x\end{array}$$

Verification

eq:=diff(y(x),x)+a*y(x)*(y(x)-x)-1 = 0; 
sol:=exp(-a*x^2/2)/(sqrt(a*Pi/2)*erf(sqrt(a/2)*x)+_C1)+x; 
odetest(y(x)=sol,eq); 
0