2.17   ODE No. 17

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

$$y^{\prime }(x)-y(x{)}^2-3y(x)+4=0$$ ✓ Mathematica : cpu = 0.0247082 (sec), leaf count = 30

$$\left\{\{y(x)\to \frac{-4e^{5(c_1+x)}-1}{e^{5(c_1+x)}-1}\}\right\}$$

✓ Maple : cpu = 0.143 (sec), leaf count = 24

$$\{y\left(x\right)=\frac{-4{\mathrm{e}}^{5x}\mathit{\_}\mathit{C}\mathit{1}-1}{-1+{\mathrm{e}}^{5x}\mathit{\_}\mathit{C}\mathit{1}}\}$$

Hand solution

$$\begin{array}{rl}{y}^{\prime }-y^2-3y+4& =0\\ \text{(1)}& y^{\prime }& =3y-4+y^2\end{array}$$

This is Riccati first order non-linear ODE of the form. The general form is$$y^{\prime }=P\left(x\right)+Q\left(x\right)y+R\left(x\right)y^2$$ Where $P\left(x\right)=-4,Q\left(x\right)=3,R\left(x\right)=1$. Using the substitution $y=-\frac{u^{\prime }}{uR\left(x\right)}=\frac{-u^{\prime }}{u}$ then$$\begin{array}{rl}{u}^{\prime }& =-yu\\ u^{\prime \prime }& =-y{u}^{\prime }-y^{\prime }u\\ & =-y(-yu)-(3y-4+y^2)u\\ & =y^{2}u-3(-\frac{u^{\prime }}{u})u+4u-y^{2}u\\ & =3u^{\prime }+4u\end{array}$$

Hence$$u^{\prime \prime }-3u^{\prime }-4u=0$$ This is standard second order ODE. The characteristic equation is  ${\lambda }^2-3\lambda -4=0$, with roots $\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{3\pm \sqrt{9+16}}{2}=\frac{3\pm 5}{2}=\{4,-1\}$, hence$$u\left(x\right)=c_1{e}^{4x}+c_2{e}^{-x}$$ And$$u^{\prime }\left(x\right)=c_{1}4e^{4x}-c_2{e}^{-x}$$ Since $y=\frac{-u^{\prime }}{u}$ then $$\begin{array}{rl}y\left(x\right)& =\frac{-c_{1}4e^{4x}+c_2{e}^{-x}}{c_1{e}^{4x}+c_2{e}^{-x}}\\ & =\frac{-\frac{c_1}{c_2}4e^{4x}+e^{-x}}{\frac{c_1}{c_2}{e}^{4x}+e^{-x}}\end{array}$$

Let $\frac{c_1}{c_2}=C_1$ then$$y\left(x\right)=\frac{-4C_1{e}^{4x}+e^{-x}}{C_1{e}^{4x}+e^{-x}}$$ Dividing by $e^{-x}$$$y\left(x\right)=\frac{-4C_1{e}^{5x}+1}{C_1{e}^{5x}+1}$$ This is the same result given by CAS. To see it better, let $C_2=-C_1$ then the above becomes$$\begin{array}{rl}y\left(x\right)& =\frac{4C_2{e}^{5x}+1}{-C_2{e}^{5x}+1}\\ & =-\frac{4C_2{e}^{5x}+1}{C_2{e}^{5x}-1}\end{array}$$