2.2.6.4.1 Example 1\(\ y^{\prime }=\left ( 1+5x+y\right ) ^{\frac {1}{2}}\) Let \(z=1+5x+y\), then \(\frac {dz}{dx}=5+y^{\prime }\). This simplifies to\begin {align*} y^{\prime } & =z^{\prime }-5\\ \left ( 1+x^{2}+y\right ) ^{\frac {1}{2}} & =z^{\prime }-5\\ z^{\frac {1}{2}} & =z^{\prime }-5\\ \frac {dz}{dx} & =z^{\frac {1}{2}}+5 \end {align*}
Which is separable. Hence \begin {align*} \frac {dz}{z^{\frac {1}{2}}+5} & =dx\qquad z^{\frac {1}{2}}+5\neq 0\\ 2\sqrt {z}-5\ln \left ( 5+\sqrt {z}\right ) +5\ln \left ( \sqrt {z}-5\right ) -5\ln \left ( z-25\right ) & =x+C_{1} \end {align*}
Hence the implicit solution is\begin {align} 2\sqrt {1+5x+y}-5\ln \left ( 5+\sqrt {1+5x+y}\right ) +5\ln \left ( \sqrt {1+5x+y}-5\right ) -5\ln \left ( 1+5x+y-25\right ) & =x+C_{1}\nonumber \\ 2\sqrt {1+5x+y}-5\ln \left ( 5+\sqrt {1+5x+y}\right ) +5\ln \left ( \sqrt {1+5x+y}-5\right ) -5\ln \left ( 5x+y-24\right ) & =x+C_{1} \tag {1} \end {align}
The above method is now compared to using d’Alembert for solving the ode, which results after squaring both sides of the given ode. Squaring the ode gives
\begin {align} \left ( y^{\prime }\right ) ^{2} & =\left ( 1+5x+y\right ) \nonumber \\ y & =\left ( y^{\prime }\right ) ^{2}-1-5x\nonumber \\ & =x\left ( -5\right ) +\left ( p^{2}-1\right ) \nonumber \\ & =xf\left ( p\right ) +g\left ( p\right ) \tag {2} \end {align}
Where \(p=\frac {dy}{dx}\). This is d’Alembert of the form \(y=xf\left ( p\right ) +g\left ( p\right ) \) where \(f\left ( p\right ) =5\) and \(g\left ( p\right ) =p^{2}-1\). Taking derivative of (2) w.r.t. \(x\) gives\begin {align} p & =f\left ( p\right ) +x\frac {df}{dp}\frac {dp}{dx}+\frac {dg}{dp}\frac {dp}{dx}\nonumber \\ p-f\left ( p\right ) & =\left ( x\frac {df}{dp}+\frac {dg}{dp}\right ) \frac {dp}{dx} \tag {3} \end {align}
Using \(f\left ( p\right ) =5\) and \(g\left ( p\right ) =p^{2}-1\) the above becomes\begin {align*} p-5 & =2p\frac {dp}{dx}\\ \frac {dp}{dx} & =\frac {p-5}{2p} \end {align*}
Which is separable. Solving for \(p\) gives\[ p=5\operatorname *{LambertW}\left ( \frac {C}{5}e^{\frac {x}{10}-1}\right ) +5 \] Substituting this back into (2) gives\begin {equation} y=-5x+\left ( \left ( 5\operatorname *{LambertW}\left ( \frac {C}{5}e^{\frac {x}{10}-1}\right ) +5\right ) ^{2}-1\right ) \tag {4} \end {equation} This is an explicit general solution for the ode \(y^{\prime }=\left ( 1+5x+y\right ) ^{\frac {1}{2}}\). The singular solution is found when \(\frac {dp}{dx}=0\) in (3) which gives\begin {align*} p-5 & =0\\ p & =5 \end {align*}
Eq (2) now becomes\begin {align} y & =-5x+\left ( 5^{2}-1\right ) \nonumber \\ & =24-5x \tag {5} \end {align}
However, and this is the problem with squaring the ode, it can be shown that both solution (4) and (5) do not verify the given \(y^{\prime }=\left ( 1+5x+y\right ) ^{\frac {1}{2}}\). What went wrong? They do verify the ode \(y^{\prime }=-\left ( 1+5x+y\right ) ^{\frac {1}{2}}\) (with minus sign). This example shows why one must be careful when squaring both sides of an ode and solving the squared version. Therefore It is better to avoid the squaring operation and to try to find a method to solve the original ode in its original form.