home
PDF (letter size)
PDF (legal size)

## Note on solving Clairaut and d’Alembert (or Lagrange) ﬁrst order ODE

October 20, 2020   Compiled on October 20, 2020 at 4:31pm

### 1 Introduction

This note is about how to solve two ODE’s, the ﬁrst is of the form \begin{equation} y\left ( x\right ) =x\frac{dy}{dx}+f\left ( \frac{dy}{dx}\right ) \tag{1} \end{equation} And the second is of the form\begin{equation} y\left ( x\right ) =xg\left ( \frac{dy}{dx}\right ) +f\left ( \frac{dy}{dx}\right ) \tag{2} \end{equation} The ﬁrst ODE above is called the Clairaut ODE and the second is called d’Alembert (also called Lagrange ODE in some books). In the above $$f$$ and $$g$$ are functions of $$p\equiv \frac{dy}{dx}$$.

The Clairaut ODE is a special case of the d’Alembert ODE when the function $$g\left ( p\right ) =p$$.

Both ODE’s are linear in $$y\left ( x\right )$$ and in $$x$$. In Clairaut ODE $$g\left ( p\right )$$ can only be $$p$$. Anything else, even $$g\left ( p\right ) =1$$, means it is not Clairaut, but can be d’Alembert.

#### 1.1 General method to solve Clairaut ODE

Starting with the Clairaut ODE as it is easier to solve than d’Alembert. Let $$p\equiv \frac{dy}{dx}$$. Eq. (1) now becomes

\begin{equation} y\left ( x\right ) =xp+f\left ( p\right ) \tag{3} \end{equation}

Taking derivative w.r.t. $$x$$ of the above gives\begin{align*} p & =p+x\frac{dp}{dx}+\frac{df\left ( p\right ) }{dp}\frac{dp}{dx}\\ 0 & =\left ( x+\frac{df\left ( p\right ) }{dp}\right ) \frac{dp}{dx} \end{align*}

Therefore either $$\frac{dp}{dx}=0$$ or $$x+\frac{\partial f\left ( p\right ) }{\partial p}=0$$. The complete integral (i.e. the general solution) is obtained from $$\frac{dp}{dx}=0$$ and the singular solution (if any) is obtained from solving $$x+\frac{df\left ( p\right ) }{dp}=0$$.

$$\frac{dp}{dx}=0$$ implies that $$p=C_{1}$$ where $$C_{1}$$ is some constant to be found. Substituting $$p=C_{1}$$ in Eq. (3) gives \begin{equation} y\left ( x\right ) =C_{1}x+f\left ( C_{1}\right ) \tag{4} \end{equation} Now the singular solution is found by solving for $$p$$ from the ODE $$x+\frac{df\left ( p\right ) }{dp}=0$$ and substituting the solution $$p$$ back into (3). This completes the solution for Clairaut ODE.

#### 1.2 General method to solve the d’Alembert ODE

Let $$p\equiv \frac{dy}{dx}$$, then Eq(2) becomes\begin{equation} y\left ( x\right ) =xg\left ( p\right ) +f\left ( p\right ) \tag{5} \end{equation} Taking derivative w.r.t. $$x$$ gives\begin{align} p & =g\left ( p\right ) +xg^{\prime }\left ( p\right ) \frac{dp}{dx}+f^{\prime }\left ( p\right ) \frac{dp}{dx}\nonumber \\ p-g\left ( p\right ) & =\left ( xg^{\prime }\left ( p\right ) +f^{\prime }\left ( p\right ) \right ) \frac{dp}{dx} \tag{6} \end{align}

When $$\frac{dp}{dx}=0$$, then $$p$$ is constant, say $$p_{0}$$. These constants are found by solving for $$p$$ from $$p-g\left ( p\right ) =0$$ and substituting the result back into (5). This gives the singular solution. The general solution is found by rewriting (6) as\begin{align*} \frac{p-g\left ( p\right ) }{xg^{\prime }\left ( p\right ) +f^{\prime }\left ( p\right ) } & =\frac{dp}{dx}\\ \frac{dx}{dp} & =\frac{xg^{\prime }\left ( p\right ) +f^{\prime }\left ( p\right ) }{p-g\left ( p\right ) }\\ \frac{dx}{dp}-x\frac{g^{\prime }\left ( p\right ) }{p-g\left ( p\right ) } & =\frac{f^{\prime }\left ( p\right ) }{p-g\left ( p\right ) } \end{align*}

$$x\left ( p\right )$$ is now the dependent variable and $$p$$ as the independent variable. The above is a linear ﬁrst order in $$x\left ( p\right )$$ since it is of the form$x^{\prime }+xG\left ( p\right ) =Q\left ( p\right )$ This is easily solved using an integration factor\begin{align*} \frac{d}{dp}\left ( xe^{\int G\left ( p\right ) dp}\right ) & =e^{\int G\left ( p\right ) dp}Q\left ( p\right ) \\ xe^{\int G\left ( p\right ) dp} & =\int e^{\int G\left ( p\right ) dp}Q\left ( p\right ) +C_{1}\\ x\left ( p\right ) & =e^{-\int G\left ( p\right ) dp}\int e^{\int G\left ( p\right ) dp}Q\left ( p\right ) +C_{1}e^{-\int G\left ( p\right ) dp} \end{align*}

Once $$x\left ( p\right )$$ is found from the above as function of $$p$$, then $$p$$ is found by inversion of the solution. Substituting this $$p$$ back into (5) gives the general solution $$y\left ( x\right )$$.

To show how these method work, the following ODE’s are now solved.

#### 1.3 General method to solve the d’Alembert ODE when f(p)=0

This is special case of the general d’Alembert $$y=xg\left ( \frac{dy}{dx}\right ) +f\left ( p\right )$$. The ODE now reduces to$y=xg\left ( \frac{dy}{dx}\right )$ Where $$f\left ( \frac{dy}{dx}\right ) =0$$ and $$g\left ( \frac{dy}{dx}\right )$$ must be non-linear in $$\frac{dy}{dx}$$. For an example, $$y\left ( x\right ) =x\left ( \frac{dy}{dx}\right ) ^{2}$$ is d’Alembert.

Let $$p\equiv \frac{dy}{dx}$$, then $$y=xg\left ( \frac{dy}{dx}\right )$$ becomes\begin{equation} y\left ( x\right ) =xg\left ( p\right ) \tag{5} \end{equation} Taking derivative w.r.t. $$x$$ gives\begin{align} p & =g\left ( p\right ) +xg^{\prime }\left ( p\right ) \frac{dp}{dx}\nonumber \\ p-g\left ( p\right ) & =xg^{\prime }\left ( p\right ) \frac{dp}{dx} \tag{6} \end{align}

When $$\frac{dp}{dx}=0$$, then $$p$$ is constant, say $$p_{0}$$. These constants are found by solving for $$p$$ from $$p-g\left ( p\right ) =0$$ and substituting the result back into (5). This gives the singular solution. The general solution is found by rewriting (6) as\begin{align*} \frac{p-g\left ( p\right ) }{xg^{\prime }\left ( p\right ) } & =\frac{dp}{dx}\\ \frac{dx}{dp} & =\frac{xg^{\prime }\left ( p\right ) }{p-g\left ( p\right ) }\\ \frac{dx}{dp}-x\frac{g^{\prime }\left ( p\right ) }{p-g\left ( p\right ) } & =0 \end{align*}

$$x\left ( p\right )$$ is now the dependent variable and $$p$$ as the independent variable. The above is a linear ﬁrst order in $$x\left ( p\right )$$ since it is of the form$x^{\prime }+xG\left ( p\right ) =0$ This is easily solved using an integration factor\begin{align*} \frac{d}{dp}\left ( xe^{\int G\left ( p\right ) dp}\right ) & =0\\ xe^{\int G\left ( p\right ) dp} & =C_{1}\\ x\left ( p\right ) & =C_{1}e^{-\int G\left ( p\right ) dp} \end{align*}

Once $$x\left ( p\right )$$ is found from the above as function of $$p$$, then $$p$$ is found by inversion of the solution. Substituting this $$p$$ back into (5) gives the general solution $$y\left ( x\right )$$.

To show how these method work, the following ODE’s are now solved.

### 2 Solved examples

 number ode transformed $$g\left ( p\right )$$ $$f\left ( p\right )$$ type $$1$$ $$x\left ( y^{\prime }\right ) ^{2}-yy^{\prime }=-1$$ $$y=xp+\frac{1}{p}$$ $$p$$ $$\frac{1}{p}$$ Clairaut $$2$$ $$y=xy^{\prime }-\left ( y^{\prime }\right ) ^{2}$$ $$y=xp-p^{2}$$ $$p$$ $$-p^{2}$$ Clairaut $$3$$ $$y=xy^{\prime }-\frac{1}{4}\left ( y^{\prime }\right ) ^{2}$$ $$y=xp-\frac{1}{4}p^{2}$$ $$p$$ $$-\frac{1}{4}p^{2}$$ Clairaut $$4$$ $$y=x\left ( y^{\prime }\right ) ^{2}$$ $$y=xp^{2}$$ $$p^{2}$$ $$0$$ d’Alembert $$5$$ $$y=x+\left ( y^{\prime }\right ) ^{2}$$ $$y=x+p^{2}$$ $$1$$ $$p^{2}$$ d’Alembert $$6$$ $$\left ( y^{\prime }\right ) ^{2}-1-x-y=0$$ $$y=-x+\left ( p^{2}-1\right )$$ $$-1$$ $$p^{2}-1$$ d’Alembert $$7$$ $$yy^{\prime }-\left ( y^{\prime }\right ) ^{2}=x$$ $$y=\frac{1}{p}x+p$$ $$\frac{1}{p}$$ $$p$$ d’Alembert $$8$$ $$y=x\left ( y^{\prime }\right ) ^{2}+\left ( y^{\prime }\right ) ^{2}$$ $$y=xp^{2}+p^{2}$$ $$p^{2}$$ $$p^{2}$$ d’Alembert $$9$$ $$y=\frac{x}{a}y^{\prime }+\frac{b}{ay^{\prime }}$$ $$y=\frac{x}{a}p+\frac{b}{a}p^{-1}$$ $$\frac{p}{a}$$ $$\frac{b}{ap}$$ d’Alembert $$10$$ $$y=x\left ( y^{\prime }+a\sqrt{1+\left ( y^{\prime }\right ) ^{2}}\right )$$ $$y=x\left ( p+a\sqrt{1+p^{2}}\right )$$ $$p+a\sqrt{1+p^{2}}$$ $$0$$ d’Alembert $$11$$ $$y=x\left ( y^{\prime }\right ) ^{2}$$ $$y=xp^{2}$$ $$p^{2}$$ $$0$$ d’Alembert

Notice in the above table, that the ODE is Clairaut only when $$g\left ( p\right ) =p$$ and d’Alembert otherwise.

#### 2.1 Example 1

To solve $$x\left ( y^{\prime }\right ) ^{2}-yy^{\prime }=-1$$, we ﬁrst write it in normal form (by replacing $$y^{\prime }$$ with $$p$$), gives\begin{equation} y\left ( x\right ) =xp+\frac{1}{p} \tag{1} \end{equation} Taking derivative w.r.t. $$x$$ gives\begin{align*} p & =p+x\frac{dp}{dx}-\frac{1}{p^{2}}\frac{dp}{dx}\\ 0 & =\left ( x-\frac{1}{p^{2}}\right ) \frac{dp}{dx} \end{align*}

The general solution is found from $$\frac{dp}{dx}=0$$. This implies $$p=C_{1}$$. Substituting this into (1) gives\begin{equation} y\left ( x\right ) =xC_{1}+\frac{1}{C_{1}} \tag{2} \end{equation} The singular solution is found by solving $$x-\frac{1}{p^{2}}=0$$. Hence $$p^{2}=\frac{1}{x}$$ or $$p=\pm \sqrt{\frac{1}{x}}$$. Substituting these back in (1) gives\begin{align} y_{1}\left ( x\right ) & =x\sqrt{\frac{1}{x}}+\sqrt{x}=2\sqrt{x}\tag{3}\\ y_{2}\left ( x\right ) & =-x\sqrt{\frac{1}{x}}-\sqrt{x}=-2\sqrt{x} \tag{4} \end{align}

Eq. (2) is the complete integral and (3,4) are the singular solutions.

#### 2.2 Example 2

To solve $$y=xy^{\prime }-\left ( y^{\prime }\right ) ^{2}$$, we ﬁrst write it in normal form (by replacing $$y^{\prime }$$ with $$p$$), gives\begin{equation} y=xp-p^{2} \tag{1} \end{equation} Taking derivative w.r.t. $$x$$ gives\begin{align*} p & =p+x\frac{dp}{dx}-2p\frac{dp}{dx}\\ 0 & =\left ( x-2p\right ) \frac{dp}{dx} \end{align*}

The general solution is found from $$\frac{dp}{dx}=0$$. This implies $$p=C_{1}$$. Substituting this into (1) gives\begin{equation} y\left ( x\right ) =xC_{1}-C_{1}^{2} \tag{2} \end{equation} The singular solution is found by solving $$x-2p=0$$. Hence $$p=\frac{x}{2}$$. Substituting these back in (1) gives\begin{align} y_{1}\left ( x\right ) & =\frac{x^{2}}{2}-\frac{x^{2}}{4}\tag{3}\\ & =\frac{x^{2}}{4}\nonumber \end{align}

Eq. (2) is the complete integral and (3) is the singular solution.

#### 2.3 Example 3

To solve $$y=xy^{\prime }-\frac{1}{4}\left ( y^{\prime }\right ) ^{2}$$, we ﬁrst write it in normal form (by replacing $$y^{\prime }$$ with $$p$$),gives\begin{equation} y=xp-\frac{1}{4}p^{2} \tag{1} \end{equation} Taking derivative w.r.t. $$x$$ gives\begin{align*} p & =p+x\frac{dp}{dx}-\frac{1}{2}p\frac{dp}{dx}\\ 0 & =\left ( x-\frac{1}{2}p\right ) \frac{dp}{dx} \end{align*}

The general solution is found from $$\frac{dp}{dx}=0$$. This implies $$p=C_{1}$$. Substituting this into (1) gives\begin{equation} y\left ( x\right ) =xC_{1}-\frac{1}{4}C_{1}^{2} \tag{2} \end{equation} The singular solution is found by solving $$x-\frac{1}{2}p=0$$. Hence $$p=2x$$. Substituting this back in (1) gives\begin{equation} y_{1}\left ( x\right ) =2x^{2}-x^{2}=x^{2} \tag{3} \end{equation} Eq. (2) is the complete integral and (3) is the singular solutions.

#### 2.4 Example 4

To solve $$y=x\left ( y^{\prime }\right ) ^{2}$$, we ﬁrst write it in normal form (by replacing $$y^{\prime }$$ with $$p$$), gives\begin{equation} y=xp^{2} \tag{1} \end{equation} Taking derivative w.r.t. $$x$$ gives\begin{align} p & =p^{2}+2xp\frac{dp}{dx}\nonumber \\ p-p^{2} & =2xp\frac{dp}{dx} \tag{2} \end{align}

To ﬁnd the singular solution, we consider when $$\frac{dp}{dx}=0,$$ which implies that $$p$$ is constant. Hence $$p-p^{2}=0$$ in this case. Solving for this gives $$p=0$$ or $$p=1$$. For each one of these values, we obtain a singular solution by substituting these values in (1), which gives\begin{align*} y_{1}\left ( x\right ) & =0\\ y_{2}\left ( x\right ) & =x \end{align*}

Now we need to ﬁnd the general solution which is when $$\frac{dp}{dx}\neq 0$$. From (2), writing it as\begin{align*} \frac{p-p^{2}}{2xp} & =\frac{dp}{dx}\\ \frac{dx}{dp} & =\frac{2xp}{p-p^{2}} \end{align*}

This is linear ODE in $$x\left ( p\right )$$. Solving it gives $$x=\frac{C_{1}}{\left ( p-1\right ) ^{2}}$$. This implies $$\left ( p-1\right ) ^{2}=\frac{C_{0}}{x}$$ or $$p-1=\pm \sqrt{\frac{C_{0}}{x}}$$ or \begin{align*} p & =1+\sqrt{\frac{C_{0}}{x}}\\ p & =1-\sqrt{\frac{C_{0}}{x}} \end{align*}

For each $$p$$, there is a solution. Substituting the above in (1) gives\begin{align*} y_{3}\left ( x\right ) & =x\left ( 1+\sqrt{\frac{C_{0}}{x}}\right ) ^{2}\\ y_{4}\left ( x\right ) & =x\left ( 1-\sqrt{\frac{C_{0}}{x}}\right ) ^{2} \end{align*}

Note however, that $$y_{2}=x$$ can be obtained from $$y_{3}\left ( x\right )$$ when $$C_{0}=0$$. Hence $$y_{2}\left ( x\right ) =x$$ is not singular solution. Therefore the ﬁnal solution is\begin{align*} y\left ( x\right ) & =0\\ y\left ( x\right ) & =x\left ( 1+\sqrt{\frac{C_{0}}{x}}\right ) ^{2}\\ y\left ( x\right ) & =x\left ( 1-\sqrt{\frac{C_{0}}{x}}\right ) ^{2} \end{align*}

#### 2.5 Example 5

To solve $$y=x+\left ( y^{\prime }\right ) ^{2}$$, we ﬁrst write it in normal form (by replacing $$y^{\prime }$$ with $$p$$), gives\begin{equation} y=x+p^{2} \tag{1} \end{equation} Taking derivative w.r.t. $$x$$ gives\begin{align} p & =1+2p\frac{dp}{dx}\nonumber \\ p-1 & =2p\frac{dp}{dx} \tag{2} \end{align}

To ﬁnd the singular solution, we consider when $$\frac{dp}{dx}=0,$$ which implies that $$p$$ is constant. Hence $$p-1=0$$ in this case. Solving for this gives $$p=1$$. Substituting this values in (1), gives the solution\begin{equation} y\left ( x\right ) =x+1 \tag{3} \end{equation} Now we need to ﬁnd the general solution which is when $$\frac{dp}{dx}\neq 0$$. From (2), writing it as\begin{align*} \frac{p-1}{2p} & =\frac{dp}{dx}\\ \frac{dx}{dp} & =\frac{2p}{p-1} \end{align*}

Integrating gives\begin{align*} x & =2p+2\ln \left ( p-1\right ) +C\\ p & =\operatorname *{LambertW}\left ( e^{\frac{x}{2}-1-\frac{C}{2}}\right ) +1\\ & =\operatorname *{LambertW}\left ( C_{1}e^{\frac{x}{2}-1}\right ) +1 \end{align*}

Substituting the above in (1) gives the general solution\begin{equation} y\left ( x\right ) =x+\left ( \operatorname *{LambertW}\left ( C_{1}e^{\frac{x}{2}-1}\right ) +1\right ) ^{2} \tag{4} \end{equation} Note however that when $$C_{1}=0$$ then the general solution becomes $$y\left ( x\right ) =x+1$$. Hence (3) is a particular solution and not a singular solution. Hence (4) is the only solution.

#### 2.6 Example 6

To solve $$\left ( y^{\prime }\right ) ^{2}-1-x-y=0$$, we ﬁrst write it in normal form (by replacing $$y^{\prime }$$ with $$p$$), gives\begin{equation} y=-x+\left ( p^{2}-1\right ) \tag{1} \end{equation} Taking derivative w.r.t. $$x$$ gives\begin{align} p & =-1+2p\frac{dp}{dx}\nonumber \\ p+1 & =2p\frac{dp}{dx} \tag{2} \end{align}

To ﬁnd the singular solution, we consider when $$\frac{dp}{dx}=0,$$ which implies that $$p$$ is constant. Hence $$p+1=0$$ in this case. Solving for this gives $$p=-1$$. Substituting this values in (1), gives the solution\begin{equation} y\left ( x\right ) =-x \tag{3} \end{equation} Now we need to ﬁnd the general solution which is when $$\frac{dp}{dx}\neq 0$$. From (2), writing it as\begin{align*} \frac{p+1}{2p} & =\frac{dp}{dx}\\ \frac{dx}{dp} & =\frac{2p}{p+1} \end{align*}

Integrating gives\begin{align*} x & =2p-2\ln \left ( p+1\right ) +C\\ p & =-\operatorname *{LambertW}\left ( -e^{-\frac{x}{2}-1+\frac{C}{2}}\right ) -1\\ & =-\operatorname *{LambertW}\left ( -C_{1}e^{-\frac{x}{2}-1}\right ) +1 \end{align*}

Substituting the above in (1) gives the general solution\begin{align} y\left ( x\right ) & =-x+\left ( p^{2}-1\right ) \nonumber \\ y\left ( x\right ) & =-x+\left ( -\operatorname *{LambertW}\left ( -C_{1}e^{-\frac{x}{2}-1}\right ) +1\right ) ^{2}-1 \tag{4} \end{align}

Note however that when $$C_{1}=0$$ then the general solution becomes $$y\left ( x\right ) =-x$$. Hence (3) is a particular solution and not a singular solution. Solution (4) is the only solution.

#### 2.7 Example 7

Solving $$yy^{\prime }-\left ( y^{\prime }\right ) ^{2}=x$$. Writing it in normal form (by replacing $$y^{\prime }$$ with $$p$$), gives\begin{equation} y=\frac{1}{p}x+p \tag{1} \end{equation} Taking derivative w.r.t. $$x$$ gives\begin{align} p & =\frac{1}{p}-\frac{x}{p^{2}}\frac{dp}{dx}+\frac{dp}{dx}\nonumber \\ p-\frac{1}{p} & =\left ( 1-\frac{x}{p^{2}}\right ) \frac{dp}{dx} \tag{2} \end{align}

The singular solution is found by setting $$\frac{dp}{dx}=0,$$ which implies that $$p$$ is a constant. In this case $$p-\frac{1}{p}=0$$. Solving for this gives $$p=\pm 1$$. Substituting these values in (1) gives the solutions\begin{align} y_{1}\left ( x\right ) & =x+1\tag{3}\\ y_{2}\left ( x\right ) & =-\left ( x+1\right ) \tag{4} \end{align}

The general solution is found when $$\frac{dp}{dx}\neq 0$$. Writing (2) as\begin{align*} \frac{\frac{p^{2}-1}{p}}{1-\frac{x}{p^{2}}} & =\frac{dp}{dx}\\ \frac{dx}{dp} & =\frac{\frac{p^{2}-x}{p^{2}}}{\frac{p^{2}-1}{p}}\\ & =\frac{p^{2}-x}{p\left ( p^{2}-1\right ) }\\ \frac{dx}{dp}+x\frac{1}{p\left ( p^{2}-1\right ) } & =\frac{p}{\left ( p^{2}-1\right ) } \end{align*}

This is linear ODE in $$x\left ( p\right )$$. The solution is\begin{align} x\left ( p\right ) & =\frac{p\sqrt{\left ( p-1\right ) \left ( 1+p\right ) }\ln \left ( p+\sqrt{p^{2}-1}\right ) }{\left ( 1+p\right ) \left ( p-1\right ) }+c_{1}\frac{p}{\sqrt{\left ( 1+p\right ) \left ( p-1\right ) }}\nonumber \\ & =\frac{p\sqrt{p^{2}-1}\ln \left ( p+\sqrt{p^{2}-1}\right ) }{p^{2}-1}+c_{1}\frac{p}{\sqrt{p^{2}-1}} \tag{5} \end{align}

From (1) since $$y=\frac{1}{p}x+p$$ then solving for $$p$$ gives\begin{align*} p_{1} & =\frac{y}{2}+\frac{1}{2}\sqrt{y^{2}-4x}\\ p_{2} & =\frac{y}{2}-\frac{1}{2}\sqrt{y^{2}-4x} \end{align*}

For each $$p_{i}$$ above, substituting into (5) gives solution (implicit) for $$y\left ( x\right )$$. First solution is$x=\frac{\left ( \frac{y}{2}+\frac{1}{2}\sqrt{y^{2}-4x}\right ) \sqrt{\left ( \frac{y}{2}+\frac{1}{2}\sqrt{y^{2}-4x}\right ) ^{2}-1}\ln \left ( \frac{y}{2}+\frac{1}{2}\sqrt{y^{2}-4x}+\sqrt{\left ( \frac{y}{2}+\frac{1}{2}\sqrt{y^{2}-4x}\right ) ^{2}-1}\right ) }{\left ( \frac{y}{2}+\frac{1}{2}\sqrt{y^{2}-4x}\right ) ^{2}-1}+c_{1}\frac{\frac{y}{2}+\frac{1}{2}\sqrt{y^{2}-4x}}{\sqrt{\left ( \frac{y}{2}+\frac{1}{2}\sqrt{y^{2}-4x}\right ) ^{2}-1}}$ And second solution is$x=\frac{\left ( \frac{y}{2}-\frac{1}{2}\sqrt{y^{2}-4x}\right ) \sqrt{\left ( \frac{y}{2}-\frac{1}{2}\sqrt{y^{2}-4x}\right ) ^{2}-1}\ln \left ( \frac{y}{2}-\frac{1}{2}\sqrt{y^{2}-4x}+\sqrt{\left ( \frac{y}{2}-\frac{1}{2}\sqrt{y^{2}-4x}\right ) ^{2}-1}\right ) }{\left ( \frac{y}{2}-\frac{1}{2}\sqrt{y^{2}-4x}\right ) ^{2}-1}+c_{1}\frac{\frac{y}{2}-\frac{1}{2}\sqrt{y^{2}-4x}}{\sqrt{\left ( \frac{y}{2}-\frac{1}{2}\sqrt{y^{2}-4x}\right ) ^{2}-1}}$

#### 2.8 Example 8

Solving $$y=x\left ( y^{\prime }\right ) ^{2}+\left ( y^{\prime }\right ) ^{2}$$. Writing it in normal form (by replacing $$y^{\prime }$$ with $$p$$), gives\begin{equation} y=xp^{2}+p^{2} \tag{1} \end{equation} Taking derivative w.r.t. $$x$$ gives\begin{align} p & =p^{2}+2xp\frac{dp}{dx}+2p\frac{dp}{dx}\nonumber \\ p-p^{2} & =\left ( 2xp+2p\right ) \frac{dp}{dx} \tag{2} \end{align}

The singular solution is found by setting $$\frac{dp}{dx}=0,$$ which implies that $$p$$ is a constant. In this case $$p\left ( 1-p\right ) =0$$. Solving for this gives $$p=0,p=1$$. Substituting these values in (1) gives the solutions\begin{align} y_{1}\left ( x\right ) & =0\tag{3}\\ y_{2}\left ( x\right ) & =x+1 \tag{4} \end{align}

The general solution is found when $$\frac{dp}{dx}\neq 0$$. Writing (2) as\begin{align*} \frac{p\left ( 1-p\right ) }{2p\left ( x+1\right ) } & =\frac{dp}{dx}\\ \frac{dx}{dp} & =\frac{2\left ( x+1\right ) }{\left ( 1-p\right ) }\\ \frac{dx}{dp}-x\frac{2}{\left ( 1-p\right ) } & =\frac{2}{\left ( 1-p\right ) } \end{align*}

This is linear ODE in $$x\left ( p\right )$$. The solution is$x=\frac{C^{2}}{\left ( p-1\right ) ^{2}}-1$ Hence\begin{align*} \frac{C^{2}}{\left ( p-1\right ) ^{2}} & =x+1\\ \left ( p-1\right ) ^{2} & =\frac{C^{2}}{x+1}\\ \left ( p-1\right ) & =\pm \frac{C}{\sqrt{x+1}}\\ p & =1\pm \frac{C}{\sqrt{x+1}} \end{align*}

Substituting the above in (1) gives the general solutions$y=\left ( x+1\right ) p^{2}$ Therefore\begin{align*} y\left ( x\right ) & =\left ( x+1\right ) \left ( 1+\frac{C}{\sqrt{x+1}}\right ) ^{2}\\ y\left ( x\right ) & =\left ( x+1\right ) \left ( 1-\frac{C}{\sqrt{x+1}}\right ) ^{2} \end{align*}

The solution $$y_{1}\left ( x\right ) =0$$ found earlier is seen to be singular since it can not be obtained from the above general solution. But $$y_{2}\left ( x\right ) =x+1$$ can be obtained from the general solution when $$C=0$$. Hence there are three solutions, they are\begin{align*} y_{1}\left ( x\right ) & =0\\ y_{2}\left ( x\right ) & =\left ( x+1\right ) \left ( 1+\frac{C}{\sqrt{x+1}}\right ) ^{2}\\ y_{3}\left ( x\right ) & =\left ( x+1\right ) \left ( 1-\frac{C}{\sqrt{x+1}}\right ) ^{2} \end{align*}

#### 2.9 Example 9

Solving $$y=\frac{x}{a}y^{\prime }+\frac{b}{ay^{\prime }}$$. Writing it in normal form (by replacing $$y^{\prime }$$ with $$p$$), gives\begin{equation} y=\frac{x}{a}p+\frac{b}{a}p^{-1}\tag{1} \end{equation} Taking derivative w.r.t. $$x$$ gives\begin{align} p & =\frac{p}{a}+\frac{x}{a}\frac{dp}{dx}-\frac{b}{a}p^{-2}\frac{dp}{dx}\nonumber \\ p-\frac{p}{a} & =\left ( \frac{x}{a}-\frac{b}{a}p^{-2}\right ) \frac{dp}{dx}\tag{2} \end{align}

The singular solution is found by setting $$\frac{dp}{dx}=0,$$ which implies that $$p$$ is a constant. In this case $$p\left ( 1-\frac{1}{a}\right ) =0$$. Solving for this gives $$p=0$$. Substituting these values in (1) does not generate any solutions due to division by zero. Hence no singular solution exist. The general solution is found when $$\frac{dp}{dx}\neq 0$$. Writing (2) as\begin{align*} \frac{p\left ( 1-\frac{1}{a}\right ) }{\frac{x}{a}-\frac{b}{a}p^{-2}} & =\frac{dp}{dx}\\ \frac{dx}{dp} & =\frac{\frac{x}{a}-\frac{b}{a}p^{-2}}{p\left ( 1-\frac{1}{a}\right ) }\\ \frac{dx}{dp}-x\frac{1}{p\left ( a-1\right ) } & =-\frac{b}{a}\frac{1}{p^{3}\left ( 1-\frac{1}{a}\right ) } \end{align*}

This is linear ODE in $$x\left ( p\right )$$. The solution is\begin{equation} x\left ( p\right ) =\frac{b}{(3a-2)p^{2}}+C_{1}p^{\frac{1}{a-1}}\tag{3} \end{equation} Solving for $$p\,$$ from above gives (using the computer) the following root object$p=\operatorname *{RootOf}\left ( 2\_Z^{\frac{1}{a-1}}C_{1}\_Z^{2}a-\_Z^{\frac{1}{a-1}}C_{1}\_Z^{2}-2\_Z^{2}ax+\_Z^{2}x+b\right )$ Substituting the above back into (1) gives the solution as$y=\frac{x}{a}\operatorname *{RootOf}\left ( 2\_Z^{\frac{1}{a-1}}C_{1}\_Z^{2}a-\_Z^{\frac{1}{a-1}}C_{1}\_Z^{2}-2\_Z^{2}ax+\_Z^{2}x+b\right ) +\frac{b}{a\operatorname *{RootOf}\left ( 2\_Z^{\frac{1}{a-1}}C_{1}\_Z^{2}a-\_Z^{\frac{1}{a-1}}C_{1}\_Z^{2}-2\_Z^{2}ax+\_Z^{2}x+b\right ) }$ The hardest step in this method is the inversion of the solution of (3) to obtain $$p$$. The above solution was veriﬁed using the computer and it satisﬁes the ode $$y=\frac{x}{a}y^{\prime }+\frac{b}{ay^{\prime }}$$

#### 2.10 Example 10

Solving $$y=xy^{\prime }+ax\sqrt{1+\left ( y^{\prime }\right ) ^{2}}$$. Writing it in normal form (by replacing $$y^{\prime }$$ with $$p$$), gives\begin{equation} y=x\left ( p+a\sqrt{1+p^{2}}\right ) \tag{1} \end{equation} Hence $$g\left ( p\right ) =p+a\sqrt{1+p^{2}},f\left ( p\right ) =0$$. This is the special case of d’Alembert when $$f\left ( p\right ) =0$$. Taking derivative w.r.t. $$x$$ gives\begin{align} p & =p+a\sqrt{1+p^{2}}+x\left ( \frac{dp}{dx}+\frac{1}{2}\frac{2ap}{\sqrt{1+p^{2}}}\frac{dp}{dx}\right ) \nonumber \\ 0 & =a\sqrt{1+p^{2}}+x\frac{dp}{dx}+\frac{1}{2}x\frac{2ap}{\sqrt{1+p^{2}}}\frac{dp}{dx}\nonumber \\ 0 & =a\sqrt{1+p^{2}}+\left ( x+\frac{1}{2}x\frac{2ap}{\sqrt{1+p^{2}}}\right ) \frac{dp}{dx}\nonumber \\ -a\sqrt{1+p^{2}} & =\left ( x+\frac{1}{2}x\frac{2ap}{\sqrt{1+p^{2}}}\right ) \frac{dp}{dx} \tag{2} \end{align}

The singular solution is found by setting $$\frac{dp}{dx}=0,$$ which implies that $$-a\sqrt{1+p^{2}}=0$$ or $$\sqrt{1+p^{2}}=0$$. This gives no real solution for $$p.$$ Hence no singular solution.

The general solution is when $$\frac{dp}{dx}\neq 0$$. Writing (2) as\begin{align*} \frac{-a\sqrt{1+p^{2}}}{x+\frac{1}{2}x\frac{2ap}{\sqrt{1+p^{2}}}} & =\frac{dp}{dx}\\ \frac{dx}{dp} & =\frac{x\left ( 1+\frac{1}{2}\frac{2ap}{\sqrt{1+p^{2}}}\right ) }{-a\sqrt{1+p^{2}}}\\ \frac{dx}{x} & =\frac{1+\frac{1}{2}\frac{2ap}{\sqrt{1+p^{2}}}}{-a\sqrt{1+p^{2}}}dp\\ \frac{dx}{x} & =\frac{\sqrt{1+p^{2}}+\frac{1}{2}2ap}{-a\left ( 1+p^{2}\right ) }dp\\ \frac{dx}{x} & =\left ( -\frac{1}{a\sqrt{1+p^{2}}}-\frac{p}{\left ( 1+p^{2}\right ) }\right ) dp \end{align*}

Integrating gives$\ln x\left ( p\right ) =-\frac{1}{2}\ln \left ( p^{2}+1\right ) -\frac{1}{a}\operatorname{arcsinh}\left ( p\right )$ Hence\begin{equation} x=c_{1}\frac{-e^{-\frac{1}{a}\left ( \operatorname{arcsinh}\left ( p\right ) \right ) }}{\sqrt{p^{2}+1}} \tag{3} \end{equation} But from (1), we see that, since $$y=x\left ( p+a\sqrt{1+p^{2}}\right )$$ then solving for $$p$$ from this gives solutions \begin{align*} p_{1} & =-\frac{1}{x}\frac{ay+\sqrt{-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\\ p_{2} & =\frac{1}{x}\frac{-ay+\sqrt{-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1} \end{align*}

We substitute back each one of the above solutions $$p_{i}$$ into Eq (3) to obtain two implicit solutions for $$y\left ( x\right )$$. The ﬁrst solution is when $$p=p_{1}$$$x=c_{1}\frac{-e^{-\frac{1}{a}\left ( \operatorname{arcsinh}\left ( -\frac{1}{x}\frac{ay+\sqrt{-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) \right ) }}{\sqrt{\left ( -\frac{1}{x}\frac{ay+\sqrt{-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) ^{2}+1}}$ And the second solution when $$p=p_{2}$$ is$x=c_{1}\frac{-e^{-\frac{1}{a}\left ( \operatorname{arcsinh}\left ( \frac{1}{x}\frac{-ay+\sqrt{-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) \right ) }}{\sqrt{\left ( \frac{1}{x}\frac{-ay+\sqrt{-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) ^{2}+1}}$

#### 2.11 Example 11

Solving $$y=x\left ( y^{\prime }\right ) ^{2}$$. Writing it in normal form (by replacing $$y^{\prime }$$ with $$p$$), gives\begin{equation} y=xp^{2} \tag{1} \end{equation} Hence $$g\left ( p\right ) =p^{2},f\left ( p\right ) =0$$. This is the special case of d’Alembert when $$f\left ( p\right ) =0$$. Taking derivative w.r.t. $$x$$ gives\begin{align} p & =p^{2}+2xp\frac{dp}{dx}\nonumber \\ p-p^{2} & =2xp\frac{dp}{dx} \tag{2} \end{align}

The singular solution is found by setting $$\frac{dp}{dx}=0,$$ which implies that $$p-p^{2}=0$$ or $$p\left ( 1-p^{2}\right ) =0$$. This gives $$p=0$$ or $$p^{2}=1$$. The ﬁrst gives $$y=0$$ and the second gives $$p=1$$ or $$p=-1$$. Therefore $$y=x$$ or $$y=-x$$. But this second solution does not satisfy the ODE. Hence only $$y=0$$ and $$y=x$$ are singular solutions.

The general solution is when $$\frac{dp}{dx}\neq 0$$. Writing (2) as\begin{align*} \frac{p-p^{2}}{2xp} & =\frac{dp}{dx}\\ \frac{dx}{dp} & =\frac{2xp}{-p-p^{2}}\\ \frac{dx}{x} & =\frac{2p}{-p-p^{2}}dp\\ & =\frac{-2}{1+p}dp \end{align*}

Integrating gives$\ln x\left ( p\right ) =-2\ln \left ( 1+p\right ) +c$ Hence\begin{align} x & =c_{1}\frac{1}{\left ( 1+p\right ) ^{2}}\nonumber \\ \left ( 1+p\right ) ^{2} & =c_{1}\frac{1}{x}\nonumber \\ 1+p & =\pm c_{1}\sqrt{\frac{1}{x}}\nonumber \\ p & =\left ( \pm c_{1}\sqrt{\frac{1}{x}}\right ) -1 \tag{3} \end{align}

From (1) $$y=xp^{2}$$ therefore the general solution is\begin{align*} y_{1} & =x\left ( c_{1}\sqrt{\frac{1}{x}}-1\right ) ^{2}=x\left ( \frac{1}{x}c_{1}^{2}-2c_{1}\sqrt{\frac{1}{x}}+1\right ) =c_{1}^{2}-2xc_{1}\sqrt{\frac{1}{x}}+x\\ y_{2} & =x\left ( -c_{1}\sqrt{\frac{1}{x}}-1\right ) ^{2}=x\left ( \frac{1}{x}c_{1}^{2}+2c_{1}\sqrt{\frac{1}{x}}+1\right ) =c_{1}^{2}+2xc_{1}\sqrt{\frac{1}{x}}+x \end{align*}

Therefore the solutions are\begin{align*} y_{1}\left ( x\right ) & =c_{1}^{2}-2xc_{1}\sqrt{\frac{1}{x}}+x\\ y_{2}\left ( x\right ) & =c_{1}^{2}+2xc_{1}\sqrt{\frac{1}{x}}+x\\ y_{3}\left ( x\right ) & =x\\ y_{4}\left ( x\right ) & =0 \end{align*}

The last two above are singular solutions.

#### 2.12 references

1. Applied diﬀerential equations, N Curle. 1972
2. Ordinary diﬀerential equations, LB Jones. 1976.
3. Elementary diﬀerential equations, William Martin, Eric Reissner. second edition. 1961.
4. Diﬀerentialgleichungen, by E. Kamke, page 30.
5. Diﬀerential and integral calculus by N. Piskunov, Vol II