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## Note on solving Clairaut or dAlembert ﬁrst order ODE’s

September 11, 2018   Compiled on September 11, 2018 at 5:31pm

### 1 Introduction

This note is about solving an ODE of the form$$y\left ( x\right ) =G\left ( x,\frac{dy}{dx}\right ) \tag{1}$$ Used to solve nonlinear ﬁrst order ODE’s. Let $p=\frac{dy}{dx}$ Then (1) becomes$$y\left ( x\right ) =G\left ( x,p\right ) \tag{2}$$ Taking derivative w.r.t $$x$$ gives$y^{\prime }\left ( x\right ) =\frac{\partial G}{\partial x}+\frac{\partial G}{\partial p}\frac{dp}{dx}$ But $$y^{\prime }\left ( x\right ) =p$$ and the above becomes\begin{align} p & =\frac{\partial G}{\partial x}+\frac{\partial G}{\partial p}\frac{dp}{dx}\nonumber \\ p-\frac{\partial G}{\partial x} & =\frac{\partial G}{\partial p}\frac{dp}{dx} \tag{3} \end{align}

There are two cases to consider here. If $$\frac{\partial G}{\partial x}=p$$, then this is called Clairaut ODE. It implies the original ODE had the form $$y\left ( x\right ) =xp+f\left ( p\right )$$. Where $$G\equiv xp+f\left ( p\right )$$ as this is only possibility to get $$\frac{\partial G}{\partial x}=p$$. Therefore $$\frac{\partial G}{\partial x}=p$$ is special case. When this happens, then (3) gives $$\frac{\partial G}{\partial p}\frac{dp}{dx}=0$$ which means either $$\frac{dp}{dx}=0$$ or $$\frac{\partial G}{\partial p}=0$$. For the case when $$\frac{dp}{dx}=0$$, this implies $$p=C_{1}$$ or since $$p=\frac{dy}{dx}$$ then $$y\left ( x\right ) =C_{1}x+C_{2}$$. Comparing $$y\left ( x\right ) =C_{1}x+C_{2}$$ to $$y\left ( x\right ) =xp+f\left ( p\right )$$ then $$C_{2}=f\left ( C_{1}\right )$$. Hence the solution is $$y\left ( x\right ) =C_{1}x+f\left ( C_{1}\right )$$. The second possibility is $$\frac{\partial G}{\partial p}=0$$. This is easily solved for $$p$$ and the second solution is found from $$y\left ( x\right ) =G\left ( x,p\right )$$ directly after that.

For the case when $$\frac{\partial G}{\partial x}\neq p$$. This is now called the d’Alembert ODE. This is harder to solve than Clairaut. To solve d’Alembert, continuing from (3) and solving (3) for $$\frac{dp}{dx}$$ results in$\frac{dp}{dx}=\left ( p-\frac{\partial G}{\partial x}\right ) \frac{\partial p}{\partial G}$ (Remembering that $$\frac{\partial G}{\partial x}\neq p$$ now). Taking $$x$$ as the dependent variable and $$p$$ as the independent variable, then solving for $$\frac{dx}{dp}$$ gives\begin{align} \frac{dx}{dp} & =\frac{\partial G}{\partial p}\left ( \frac{1}{p-\frac{\partial G}{\partial x}}\right ) \nonumber \\ \frac{dx}{dp}\left ( p-\frac{\partial G}{\partial x}\right ) & =\frac{\partial G}{\partial p} \tag{4} \end{align}

This will turn out to be a linear ODE in $$x\left ( p\right )$$, where $$x$$ is the dependent variable and $$p$$ is the independent variable in (3).

This ODE is now solved for $$x\left ( p\right )$$. Then $$p$$ is solved for from this solution in terms of $$x\left ( p\right )$$. This step is the hardest part of this method. Once $$p$$ is found, then $$y\left ( x\right ) \,$$ is found by direct substitution back into (1) because $$p=\frac{dy}{dx}$$.

To show how this method works, the following ODE’s are now solved.

 number ode transformed $$G\left ( x,p\right )$$ $$\frac{\partial G}{\partial x}$$ $$\frac{\partial G}{\partial p}$$ type $$1$$ $$\left ( y^{\prime }\right ) ^{2}-1-x-y=0$$ $$y=-x+\left ( p^{2}-1\right )$$ $$-x+\left ( p^{2}-1\right )$$ $$-1$$ $$2p$$ d’Alembert $$2$$ $$yy^{\prime }-\left ( y^{\prime }\right ) ^{2}=x$$ $$y=\frac{1}{p}x+p$$ $$\frac{1}{p}x+p$$ $$\frac{1}{p}$$ $$-\frac{1}{p^{2}}+1$$ d’Alembert $$3$$ $$x\left ( y^{\prime }\right ) ^{2}-yy^{\prime }=-1$$ $$y=xp+\frac{1}{p}$$ $$xp+\frac{1}{p}$$ $$p$$ $$x-\frac{1}{p^{2}}$$ Clairaut $$4$$ $$y=x\left ( y^{\prime }\right ) ^{2}+y^{\prime }$$ $$y=xp^{2}+p$$ $$xp^{2}+p$$ $$p^{2}$$ $$2p+1$$ d’Alembert $$5$$ $$y=xy^{\prime }-\left ( y^{\prime }\right ) ^{2}$$ $$y=xp-p^{2}$$ $$xp-p^{2}$$ $$p$$ $$x-2p$$ Clairaut $$6$$ $$y=xy^{\prime }-\frac{1}{4}\left ( y^{\prime }\right ) ^{2}$$ $$y=xp-\frac{1}{4}p^{2}$$ $$xp-\frac{1}{4}p^{2}$$ $$p$$ $$x-\frac{1}{2}p$$ Clairaut $$7$$ $$y=x\left ( y^{\prime }\right ) ^{2}$$ $$y=xp^{2}$$ $$xp^{2}$$ $$p^{2}$$ $$2xp$$ d’Alembert $$8$$ $$y=x+\left ( y^{\prime }\right ) ^{2}$$ $$y=x+p^{2}$$ $$x+p^{2}$$ $$1$$ $$2p$$ d’Alembert

Notice in the above table, that the ODE is Clairaut when $$\frac{\partial G}{\partial x}=p$$ and d’Alembert otherwise.

#### 1.1 Example 1

To solve $$\left ( y^{\prime }\right ) ^{2}-1-x-y=0$$, it is ﬁrst converted to $$y\left ( x\right ) =G\left ( x,y^{\prime }\left ( x\right ) \right )$$ which gives$$y\left ( x\right ) =-x+\left ( \left ( y^{\prime }\right ) ^{2}-1\right ) \tag{1}$$ Using $$p=\frac{dy}{dx}$$ then (1) becomes \begin{align} y\left ( x\right ) & =G\left ( x,p\right ) \nonumber \\ & =-x+\left ( p^{2}-1\right ) \tag{2} \end{align}

Hence $$\frac{\partial G}{\partial x}=-1$$ and $$\frac{\partial G}{\partial p}=2p$$. Since $$\frac{\partial G}{\partial x}\neq p$$ then this is d’Alembert. Therefore equation (4) above is used to solve the ode. (4)  becomes\begin{align*} \frac{dx}{dp}\left ( p-\frac{\partial G}{\partial x}\right ) & =\frac{\partial G}{\partial p}\\ \frac{dx}{dp}\left ( p+1\right ) & =2p\\ \frac{dx}{dp} & =\frac{2p}{p+1} \end{align*}

The solution to the above ODE is $$x\left ( p\right ) =2p-2\ln (p+1)+C_{1}$$. Now $$p$$ is solved for in terms of $$x$$ (this is hard step in this algorithm). This might not always be possible and $$\operatorname *{RootOf}$$ might have to be used. But in this example, the solution is$p=-\operatorname *{LambertW}\left ( -C_{1}e^{-\frac{x}{2}-1}\right ) -1$ Substituting the above back into (2), since $$p=y^{\prime }\left ( x\right )$$, gives the solution directly$y\left ( x\right ) =-x+\left ( -\operatorname *{LambertW}\left ( -C_{1}e^{-\frac{x}{2}-1}\right ) -1\right ) ^{2}-1$

#### 1.2 Example 2

To solve $$x=yy^{\prime }-\left ( y^{\prime }\right ) ^{2}$$, it is ﬁrst converted to $$y\left ( x\right ) =G\left ( x,y^{\prime }\left ( x\right ) \right )$$ which gives\begin{align} y\left ( x\right ) & =G\left ( x,y^{\prime }\left ( x\right ) \right ) \nonumber \\ & =\frac{x+\left ( y^{\prime }\right ) ^{2}}{y^{\prime }} \tag{1} \end{align}

Using $$p=\frac{dy}{dx}$$ then (1) becomes\begin{align} y\left ( x\right ) & =G\left ( x,p\right ) \nonumber \\ & =\frac{x}{p}+p \tag{2} \end{align}

Hence $$\frac{\partial G}{\partial x}=\frac{1}{p}$$ and $$\frac{\partial G}{\partial p}=\frac{x}{p^{2}}+1$$. Since $$\frac{\partial G}{\partial x}\neq p$$ then this is d’Alembert. Therefore (4) becomes\begin{align*} \frac{dx}{dp}\left ( p-\frac{\partial G}{\partial x}\right ) & =\frac{\partial G}{\partial p}\\ \frac{dx}{dp}\left ( p-\frac{1}{p}\right ) & =\frac{x}{p^{2}}+1\\ \frac{dx}{dp}\left ( \frac{p^{2}-1}{p}\right ) & =\frac{x}{p^{2}}+1\\ \frac{dx}{dp} & =\frac{x}{p^{2}}\left ( \frac{p}{p^{2}-1}\right ) +\frac{p}{p^{2}-1}\\ \frac{dx}{dp}-\frac{x}{p\left ( p^{2}-1\right ) } & =\frac{p}{p^{2}-1} \end{align*}

This is linear in $$x\left ( p\right ) .$$ Solving the above gives$x\left ( p\right ) =\frac{p\sqrt{\left ( p-1\right ) \left ( 1+p\right ) }\ln \left ( p+\sqrt{p^{2}-1}\right ) }{\left ( 1+p\right ) \left ( p-1\right ) }+\frac{pC_{1}}{\sqrt{\left ( p-1\right ) \left ( 1+p\right ) }}$ Now $$p$$ has to be solved in terms of $$x$$ and hence solution is found from (1).

#### 1.3 Example 3

Solving $$x\left ( y^{\prime }\right ) ^{2}-yy^{\prime }=-1$$. This is written as\begin{align} y & =xy^{\prime }\left ( x\right ) +\frac{1}{y^{\prime }\left ( x\right ) }\nonumber \\ & =G\left ( x,y^{\prime }\right ) \tag{1} \end{align}

Let $$p=\frac{dy}{dx}$$ then (1) becomes\begin{align} y\left ( x\right ) & =G\left ( x,p\right ) \nonumber \\ & =xp+\frac{1}{p} \tag{2} \end{align}

Therefore $$\frac{\partial G}{\partial x}=p$$ and $$\frac{\partial G}{\partial p}=x-\frac{1}{p^{2}}$$. Since $$\frac{\partial G}{\partial x}=p$$ then this is Clairaut ODE. This is special case. From (3),$\frac{\partial G}{\partial p}\frac{dp}{dx}=0$ First possibility is  $$\frac{dp}{dx}=0$$ which gives $$y\left ( x\right ) =C_{1}x+C_{2}$$ where $$C_{2}=f\left ( C_{1}\right )$$ and $$f\left ( p\right ) =\frac{1}{p}$$ in this case looking at (1). Hence $$C_{1}=\frac{1}{C_{1}}$$. Therefore the solution is$y_{1}\left ( x\right ) =C_{1}x+\frac{1}{C_{1}}$ The other solution from considering $$\frac{\partial G}{\partial p}=0$$ or $$x-\frac{1}{p^{2}}=0$$ which implies $$p=\pm \frac{1}{\sqrt{x}}$$. Substituting this into (2) gives 2 additional solutions\begin{align*} y_{1}\left ( x\right ) & =x\frac{1}{\sqrt{x}}+\sqrt{x}=2\sqrt{x}\\ y_{2}\left ( x\right ) & =-\left ( x\frac{1}{\sqrt{x}}+\sqrt{x}\right ) =-2\sqrt{x} \end{align*}

Therefore, the three solutions are\begin{align*} y_{1}\left ( x\right ) & =C_{1}x+\frac{1}{C_{1}}\\ y_{2}\left ( x\right ) & =2\sqrt{x}\\ y_{3}\left ( x\right ) & =-2\sqrt{x} \end{align*}

The solutions $$y_{2},y_{3}$$ are singular solutions, since they can not be obtained from the general solution $$y_{1}\left ( x\right ) =C_{1}x+\frac{1}{C_{1}}$$ by giving a speciﬁc value for $$C_{1}$$.

#### 1.4 Example 4

Solving $$y=x\left ( y^{\prime }\right ) ^{2}+y^{\prime }$$. It is transformed to\begin{align} y & =G\left ( x,p\right ) \nonumber \\ & =xp^{2}+p \tag{1} \end{align}

Hence $$\frac{\partial G}{\partial x}=p^{2}$$ and $$\frac{\partial G}{\partial p}=2xp+1$$. Since $$\frac{\partial G}{\partial x}\neq p$$ then this is d’Alembert. Therefore (4) becomes\begin{align*} \frac{dx}{dp}\left ( p-\frac{\partial G}{\partial x}\right ) & =\frac{\partial G}{\partial p}\\ \frac{dx}{dp}\left ( p-p^{2}\right ) & =2xp+1\\ \frac{dx}{dp} & =\frac{2xp}{p-p^{2}}+\frac{1}{p-p^{2}}\\ \frac{dx}{dp}-x\frac{2}{1-p} & =\frac{1}{p-p^{2}} \end{align*}

This is linear in $$x\left ( p\right )$$. Its solution is$x\left ( p\right ) =\frac{-p+\ln \left ( p\right ) +C_{1}}{\left ( p-1\right ) ^{2}}$ Now $$p$$ is solved in terms of $$x$$. This gives$$p=e^{\operatorname *{RootOf}\left ( -\left ( e^{Z}\right ) ^{2}x+2e^{Z}x+Z+C_{1}-e^{Z}-x\right ) } \tag{2}$$ Hence the solution now found for $$y\left ( x\right )$$ from (1)$y\left ( x\right ) =xe^{2\operatorname *{RootOf}\left ( -\left ( e^{Z}\right ) ^{2}x+2e^{Z}x+Z+C_{1}-e^{Z}-x\right ) }+e^{\operatorname *{RootOf}\left ( -\left ( e^{Z}\right ) ^{2}x+2e^{Z}x+Z+C_{1}-e^{Z}-x\right ) }$

#### 1.5 Example 5

Solving $$y=xy^{\prime }-\left ( y^{\prime }\right ) ^{2}$$. From above, it is transformed to\begin{align} y & =G\left ( x,p\right ) \nonumber \\ & =xp-p^{2} \tag{1} \end{align}

Therefore $$\frac{\partial G}{\partial x}=p$$ and $$\frac{\partial G}{\partial p}=x-2p$$. Since $$\frac{\partial G}{\partial x}=p$$ then this is Clairaut ODE. This is special case. From (3), $\frac{\partial G}{\partial p}\frac{dp}{dx}=0$ First possibility is  $$\frac{dp}{dx}=0$$ which gives $$y\left ( x\right ) =C_{1}x+C_{2}$$ where $$C_{2}=f\left ( C_{1}\right )$$ and $$f\left ( p\right ) =-p^{2}$$ in this case looking at (1). Hence $$C_{2}=-C_{1}^{2}$$. Therefore the solution is$y_{1}\left ( x\right ) =C_{1}x-C_{1}^{2}$ The other solution from considering $$\frac{\partial G}{\partial p}=0$$ or $$x-2p=0$$, hence $$p=\frac{x}{2}$$. Therefore from (1)\begin{align*} y\left ( x\right ) & =x\left ( \frac{x}{2}\right ) -\left ( \frac{x}{2}\right ) ^{2}\\ & =\frac{1}{4}x^{2} \end{align*}

Therefore the solutions are\begin{align*} y_{1}\left ( x\right ) & =C_{1}x-C_{1}^{2}\\ y_{2}\left ( x\right ) & =\frac{1}{4}x^{2} \end{align*}

The solution $$y_{2}\left ( x\right ) =\frac{1}{4}x^{2}$$ is singular since it can not be obtained from $$y_{1}\left ( x\right ) =C_{1}x-C_{1}^{2}$$.

#### 1.6 Example 6

Solving $$y=xy^{\prime }-\frac{1}{4}\left ( y^{\prime }\right ) ^{2}$$. From above, it is transformed to\begin{align} y & =G\left ( x,p\right ) \nonumber \\ & =xp-\frac{1}{4}p^{2} \tag{1} \end{align}

Therefore $$\frac{\partial G}{\partial x}=p$$ and $$\frac{\partial G}{\partial p}=x-\frac{1}{2}p$$. Since $$\frac{\partial G}{\partial x}=p$$ then this is Clairaut ODE. This is special case. From (3),$\frac{\partial G}{\partial p}\frac{dp}{dx}=0$ First possibility is $$\frac{dp}{dx}=0$$ which gives $$y\left ( x\right ) =C_{1}x+C_{2}$$ where $$C_{2}=f\left ( C_{1}\right )$$ and $$f\left ( p\right ) =-\frac{1}{4}p^{2}$$ in this case looking at (1). Hence $$C_{2}=-\frac{1}{4}C_{1}^{2}$$. Therefore the solution is$y_{1}\left ( x\right ) =C_{1}x-\frac{1}{4}C_{1}^{2}$ The other solution from considering $$\frac{\partial G}{\partial p}=0$$ or $$x-\frac{1}{2}p=0$$, hence $$p=2x$$. Therefore from (1)\begin{align*} y\left ( x\right ) & =x\left ( 2x\right ) -\frac{1}{4}\left ( 2x\right ) ^{2}\\ & =x^{2} \end{align*}

Therefore the solutions are\begin{align*} y_{1}\left ( x\right ) & =C_{1}x-\frac{1}{4}C_{1}^{2}\\ y_{2}\left ( x\right ) & =x^{2} \end{align*}

The solution $$y_{2}\left ( x\right ) =x^{2}$$ is singular since it can not be obtained from $$y_{1}\left ( x\right ) =C_{1}x-\frac{1}{4}C_{1}^{2}$$.

#### 1.7 Example 7

Solving $$y=x\left ( y^{\prime }\right ) ^{2}$$. From above, it is transformed to\begin{align} y & =G\left ( x,p\right ) \nonumber \\ & =xp^{2} \tag{1} \end{align}

Hence $$\frac{\partial G}{\partial x}=p^{2}$$ and $$\frac{\partial G}{\partial p}=2px$$. Since $$\frac{\partial G}{\partial x}\neq p$$ then this is d’Alembert. Therefore (4) becomes\begin{align*} \frac{dx}{dp}\left ( p-\frac{\partial G}{\partial x}\right ) & =\frac{\partial G}{\partial p}\\ \frac{dx}{dp}\left ( p-p^{2}\right ) & =2xp\\ \frac{dx}{dp} & =\frac{2x}{1-p^{2}}\\ \frac{dx}{dp}-x\frac{2}{1-p} & =0 \end{align*}

This is linear in $$x\left ( p\right )$$. Solving for $$x\left ( p\right )$$ gives$x\left ( p\right ) =\frac{C_{1}}{\left ( p-1\right ) ^{2}}$ Hence solving for $$p$$ gives\begin{align*} p_{1} & =\frac{x+\sqrt{C_{1}x}}{x}\\ p_{2} & =\frac{-x+\sqrt{C_{1}x}}{x} \end{align*}

From (1) the solutions are\begin{align*} y_{1}\left ( x\right ) & =xp_{1}^{2}\\ & =x\left ( \frac{x+\sqrt{C_{1}x}}{x}\right ) ^{2}\\ & =x+C_{1}+2\sqrt{xC_{1}} \end{align*}

And\begin{align*} y_{2}\left ( x\right ) & =xp_{1}^{2}\\ & =x\left ( \frac{-x+\sqrt{C_{1}x}}{x}\right ) ^{2}\\ & =x+C_{1}-2\sqrt{xC_{1}} \end{align*}

#### 1.8 Example 8

Solving $$y=x+\left ( y^{\prime }\right ) ^{2}$$. It is transformed to\begin{align} y & =G\left ( x,p\right ) \nonumber \\ & =x+p^{2} \tag{1} \end{align}

Hence $$\frac{\partial G}{\partial x}=1$$ and $$\frac{\partial G}{\partial p}=2p$$. Since $$\frac{\partial G}{\partial x}\neq p$$ then this is d’Alembert. Therefore (4) becomes\begin{align*} \frac{dx}{dp}\left ( p-\frac{\partial G}{\partial x}\right ) & =\frac{\partial G}{\partial p}\\ \frac{dx}{dp}\left ( p-1\right ) & =2p\\ \frac{dx}{dp} & =\frac{2p}{p-1} \end{align*}

Hence $$x=2p+2\ln \left ( p-1\right ) +C_{1}$$.  Solving for $$p$$ in terms of $$x$$ gives $$p=\operatorname *{LambertW}\left ( C_{1}e^{\frac{x}{2}-1}\right ) +1$$.  Substituting this in (1) gives the solution$y\left ( x\right ) =x+\left ( \operatorname *{LambertW}\left ( C_{1}e^{\frac{x}{2}-1}\right ) +1\right ) ^{2}$

#### 1.9 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.