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Note on solving Clairaut or dAlembert first order ODE’s

Nasser M. Abbasi

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

1 Introduction

This note is about solving an ODE of the form\begin{equation} y\left ( x\right ) =G\left ( x,\frac{dy}{dx}\right ) \tag{1} \end{equation} Used to solve nonlinear first order ODE’s. Let \[ p=\frac{dy}{dx}\] Then (1) becomes\begin{equation} y\left ( x\right ) =G\left ( x,p\right ) \tag{2} \end{equation} 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\begin{equation} y\left ( x\right ) =-x+\left ( \left ( y^{\prime }\right ) ^{2}-1\right ) \tag{1} \end{equation} 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\begin{equation} p=e^{\operatorname *{RootOf}\left ( -\left ( e^{Z}\right ) ^{2}x+2e^{Z}x+Z+C_{1}-e^{Z}-x\right ) } \tag{2} \end{equation} 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.