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Note on solving Clairaut or dAlembert first order ODE’s

Nasser M. Abbasi

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

pict

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

pict

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

pict

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

pict

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

pict

Using     dy
p = dx  then (1) becomes

pict

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

pict

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

pict

Let     dy
p = dx  then (1) becomes

pict

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

pict

Therefore, the three solutions are

pict

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

pict

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

pict

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

pict

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)

pict

Therefore the solutions are

pict

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

pict

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)

pict

Therefore the solutions are

pict

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

pict

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

pict

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

x (p ) =---C1--2
       (p − 1)

Hence solving for p  gives

pict

From (1) the solutions are

pict

And

pict

1.8 Example 8

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

pict

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

pict

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

  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.