#### 2.17   ODE No. 17

$y'(x)-y(x)^2-3 y(x)+4=0$ Mathematica : cpu = 0.0525925 (sec), leaf count = 34

$\left \{\left \{y(x)\to \frac {-4 e^{5 c_1+5 x}-1}{e^{5 c_1+5 x}-1}\right \}\right \}$ Maple : cpu = 0.092 (sec), leaf count = 24

$\left \{ y \left ( x \right ) ={\frac {-4\,{{\rm e}^{5\,x}}{\it \_C1}-1}{-1+{{\rm e}^{5\,x}}{\it \_C1}}} \right \}$

Hand solution

\begin {align} y^{\prime }-y^{2}-3y+4 & =0\nonumber \\ y^{\prime } & =3y-4+y^{2}\tag {1} \end {align}

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 {align*} u^{\prime } & =-yu\\ u^{\prime \prime } & =-yu^{\prime }-y^{\prime }u\\ & =-y\left ( -yu\right ) -\left ( 3y-4+y^{2}\right ) u\\ & =y^{2}u-3\left ( -\frac {u^{\prime }}{u}\right ) u+4u-y^{2}u\\ & =3u^{\prime }+4u \end {align*}

Hence\begin {equation} u^{\prime \prime }-3u^{\prime }-4u=0\nonumber \end {equation} This is standard second order ODE. The characteristic equation is  $$\lambda ^{2}-3\lambda -4=0$$, with roots $$\frac {-b\pm \sqrt {b^{2}-4ac}}{2a}=\frac {3\pm \sqrt {9+16}}{2}=\frac {3\pm 5}{2}=\left \{ 4,-1\right \}$$, 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 {align*} 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 {align*}

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 {align*} 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 {align*}