Note on solving Clairaut or dAlembert first order ODE’s

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Note on solving Clairaut or dAlembert first order ODE’s

September 4, 2018 Compiled on September 4, 2018 at 2:21am

1 Introduction

This note is about solving an ODE of the form

( ) dy- y(x) = G x,dx

(1)

Used to solve nonlinear first order ODE’s. Let

dy- p = dx

Then (1) becomes

y(x) = G (x,p )

(2)

Taking derivative w.r.t $x$ gives

∂G ∂G dp y′(x ) =--- + --- --- ∂x ∂p dx

But $y′(x) = p$ and the above becomes

There are two cases to consider here. If $∂G- ∂x = p$ , then this is called Clairaut ODE. It implies the original ODE had the form $y(x) = xp + f (p)$ . Where $G ≡ xp+ f (p)$ as this is only possibility to get $∂G∂x-= p$ . Therefore $∂∂Gx-= p$ is special case. When this happens, then (3) gives $∂Gdp = 0 ∂pdx$ which means either $dp= 0 dx$ or $∂G-= 0 ∂p$ . For the case when $dp dx = 0$ , this implies $p = C1$ or since $dy p = dx$ then $y (x ) = C1x + C2$ . Comparing $y (x ) = C1x + C2$ to $y (x) = xp + f (p)$ then $C2 = f (C1)$ . Hence the solution is $y (x) = C1x + f (C1)$ . The second possibility is $∂G∂p-= 0$ . This is easily solved for $p$ and the second solution is found from $y (x) = G (x,p)$ directly after that.

For the case when $∂G ∂x-⁄= p$ . 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 $ddpx$ results in

( ) dp-= p− ∂G- ∂p- dx ∂x ∂G

(Remembering that $∂G-⁄= p ∂x$ now). Taking $x$ as the dependent variable and $p$ as the independent variable, then solving for $dx dp$ gives

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

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

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


number	ode	transformed	$G (x,p)$	$∂G- ∂x$	$∂G- ∂p$	type

$1$	$′ 2 (y ) − 1 − x − y = 0$	$( 2 ) y = − x+ p − 1$	$( 2 ) − x + p − 1$	$− 1$	$2p$	d’Alembert

$2$	$yy′ − (y′)2 = x$	$y = 1px + p$	$1px+ p$	$1p$	$− 1p2-+ 1$	d’Alembert

$3$	$x (y′)2 − yy ′ = − 1$	$y = xp + 1 p$	$xp + 1 p$	$p$	$x − 12 p$	Clairaut

$4$	$′2 ′ y = x (y ) + y$	$2 y = xp + p$	$2 xp + p$	$2 p$	$2p+ 1$	d’Alembert

$5$	$′ ′ 2 y = xy − (y )$	$2 y = xp − p$	$2 xp − p$	$p$	$x − 2p$	Clairaut

$6$	$1 2 y = xy′ − 4 (y′)$	$1 y = xp − 4p2$	$1 xp − 4p2$	$p$	$1 x − 2p$	Clairaut

$7$	$y = x (y ′)2$	$y = xp2$	$xp2$	$p2$	$2xp$	d’Alembert

$8$	$y = x + (y′)2$	$y = x + p2$	$x + p2$	$1$	$2p$	d’Alembert

Notice in the above table, that the ODE is Clairaut when $∂∂Gx-= p$ and d’Alembert otherwise.

1.1 Example 1

To solve $(y′)2 − 1− x − y = 0$ , it is first converted to $y (x) = G (x,y′(x))$ which gives

( ) y(x) = − x+ (y′)2 − 1

(1)

Using $p = ddyx$ then (1) becomes

Hence $∂G- ∂x = − 1$ and $∂G- ∂p = 2p$ . Since $∂G- ∂x ⁄= p$ then this is d’Alembert. Therefore equation (4) above is used to solve the ode. (4) becomes

The solution to the above ODE is $x(p) = 2p− 2 ln(p + 1)+ C1$ . Now $p$ is solved for in terms of $x$ (this is hard step in this algorithm). This might not always be possible and $RootOf$ might have to be used. But in this example, the solution is

( x ) p = − LambertW − C1e −2− 1 − 1

Substituting the above back into (2), since $p = y′(x)$ , gives the solution directly

( ( ) ) y(x) = − x + − LambertW − C e− x2− 1 − 1 2 − 1 1

1.2 Example 2

To solve $′ ′2 x = yy − (y )$ , it is first converted to $′ y(x) = G (x,y (x))$ which gives

Using $dy p = dx$ then (1) becomes

Hence $∂∂Gx-= 1p$ and $∂G∂p-= xp2 + 1$ . Since $∂∂Gx-⁄= p$ then this is d’Alembert. Therefore (4) becomes

This is linear in $x (p).$ Solving the above gives

------------- ( ------) p∘ (p− 1) (1 + p)ln p+ ∘ p2 − 1 x(p) = ---------------------------------+ ∘-----pC1------ (1 + p)(p− 1) (p− 1)(1+ p)

Now $p$ has to be solved in terms of $x$ and hence solution is found from (1).

1.3 Example 3

Solving $2 x(y′) − yy′ = − 1$ . This is written as

Let $dy p = dx$ then (1) becomes

Therefore $∂G- ∂x = p$ and $∂G- 1- ∂p = x− p2$ . Since $∂G- ∂x = p$ then this is Clairaut ODE. This is special case. From (3),

∂G-dp-= 0 ∂p dx

First possibility is $dp= 0 dx$ which gives $y(x) = C x + C 1 2$ where $C = f (C ) 2 1$ and $1 f (p) = p$ in this case looking at (1). Hence $-1 C1 = C1$ . Therefore the solution is

1 y1(x) = C1x + --- C1

The other solution from considering $∂∂Gp-= 0$ or $x− 1p2-= 0$ which implies $p = ± √1- x$ . Substituting this into (2) gives 2 additional solutions

Therefore, the three solutions are

The solutions $y2,y3$ are singular solutions, since they can not be obtained from the general solution $y1(x) = C1x + -1 C1$ by giving a specific value for $C1$ .

1.4 Example 4

Solving $′2 ′ y = x (y) + y$ . It is transformed to

Hence $∂∂Gx-= p2$ and $∂∂Gp-= 2xp + 1$ . Since $∂∂Gx-⁄= p$ then this is d’Alembert. Therefore (4) becomes

This is linear in $x (p)$ . Its solution is

− p + ln (p)+ C x (p) =-----------2--1 (p − 1)

Now $p$ is solved in terms of $x$ . This gives

( ) RootOf− (eZ)2x+2eZx+Z+C1− eZ− x p = e

(2)

Hence the solution now found for $y (x)$ from (1)

2RootOf(−(eZ)2x+2eZx+Z+C1− eZ −x) RootOf(−(eZ )2x+2eZx+Z+C1 −eZ−x) y(x) = xe + e

1.5 Example 5

Solving $y = xy ′ − (y′)2$ . From above, it is transformed to

Therefore $∂∂Gx-= p$ and $∂G∂p = x− 2p$ . Since $∂∂Gx-= p$ then this is Clairaut ODE. This is special case. From (3),

∂G-dp-= 0 ∂p dx

First possibility is $-dp = 0 dx$ which gives $y(x) = C x + C 1 2$ where $C = f (C ) 2 1$ and $2 f (p) = − p$ in this case looking at (1). Hence $2 C2 = − C1$ . Therefore the solution is

y1(x) = C1x − C21

The other solution from considering $∂G--= 0 ∂p$ or $x − 2p = 0$ , hence $p = x 2$ . Therefore from (1)

Therefore the solutions are

The solution $1 2 y2 (x ) = 4 x$ is singular since it can not be obtained from $2 y1(x) = C1x − C1$ .

1.6 Example 6

Solving $y = xy ′ − 14 (y′)2$ . From above, it is transformed to

Therefore $∂G-= p ∂x$ and $∂G-= x− 1p ∂p 2$ . Since $∂G-= p ∂x$ then this is Clairaut ODE. This is special case. From (3),

∂G dp ------= 0 ∂p dx

First possibility is $ddpx = 0$ which gives $y(x) = C1x + C2$ where $C2 = f (C1 )$ and $f (p) = − 1p2 4$ in this case looking at (1). Hence $C2 = − 1C2 4 1$ . Therefore the solution is

1 2 y1 (x ) = C1x − 4C1

The other solution from considering $∂G- ∂p = 0$ or $1 x − 2p = 0$ , hence $p = 2x$ . Therefore from (1)

Therefore the solutions are

The solution $2 y2 (x ) = x$ is singular since it can not be obtained from $1 2 y1 (x ) = C1x − 4C1$ .

1.7 Example 7

Solving $2 y = x (y′)$ . From above, it is transformed to

Hence $∂G-= p2 ∂x$ and $∂G-= 2px ∂p$ . Since $∂G-⁄= p ∂x$ then this is d’Alembert. Therefore (4) becomes

This is linear in $x (p)$ . Solving for $x(p)$ gives

x (p ) =---C1--2 (p − 1)

Hence solving for $p$ gives

From (1) the solutions are

And

1.8 Example 8

Solving $2 y = x + (y′)$ . It is transformed to

Hence $∂G-= 1 ∂x$ and $∂G-= 2p ∂p$ . Since $∂G-⁄= p ∂x$ then this is d’Alembert. Therefore (4) becomes

Hence $x = 2p + 2ln(p − 1)+ C1$ . Solving for $p$ in terms of $x$ gives $( ) p = LambertW C1ex2− 1 + 1$ . Substituting this in (1) gives the solution

( ( x ) )2 y(x) = x+ LambertW C1e 2−1 + 1

1.9 references

Applied differential equations, N Curle. 1972
Ordinary differential equations, LB Jones. 1976.
Elementary differential equations, William Martin, Eric Reissner. second edition. 1961.
Differentialgleichungen, by E. Kamke, page 30.