\[ y'(x)-y(x)^2-3 y(x)+4=0 \] ✓ Mathematica : cpu = 0.067367 (sec), leaf count = 34
\[\left \{\left \{y(x)\to \frac {-1-4 e^{5 x+5 c_1}}{-1+e^{5 x+5 c_1}}\right \}\right \}\] ✓ Maple : cpu = 0.12 (sec), leaf count = 24
\[y \relax (x ) = \frac {-4 \,{\mathrm e}^{5 x} c_{1}-1}{-1+{\mathrm e}^{5 x} c_{1}}\]
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\[ y^{\prime }=P\relax (x) +Q\relax (x) y+R\relax (x) y^{2}\] Where \(P\relax (x) =-4,Q\relax (x) =3,R\relax (x) =1\). Using the substitution \(y=-\frac {u^{\prime }}{uR\relax (x) }=\frac {-u^{\prime }}{u}\) 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\[ u\relax (x) =c_{1}e^{4x}+c_{2}e^{-x}\] And\[ u^{\prime }\relax (x) =c_{1}4e^{4x}-c_{2}e^{-x}\] Since \(y=\frac {-u^{\prime }}{u}\) then \begin {align*} y\relax (x) & =\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\[ y\relax (x) =\frac {-4C_{1}e^{4x}+e^{-x}}{C_{1}e^{4x}+e^{-x}}\] Dividing by \(e^{-x}\)\[ y\relax (x) =\frac {-4C_{1}e^{5x}+1}{C_{1}e^{5x}+1}\] This is the same result given by CAS. To see it better, let \(C_{2}=-C_{1}\) then the above becomes\begin {align*} y\relax (x) & =\frac {4C_{2}e^{5x}+1}{-C_{2}e^{5x}+1}\\ & =-\frac {4C_{2}e^{5x}+1}{C_{2}e^{5x}-1} \end {align*}