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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

 (1)

Used to solve nonlinear first order ODE’s. Let

Then (1) becomes

 (2)

Taking derivative w.r.t gives

But and the above becomes

There are two cases to consider here. If , then this is called Clairaut ODE. It implies the original ODE had the form . Where as this is only possibility to get . Therefore is special case. When this happens, then (3) gives which means either or . For the case when , this implies or since then . Comparing to then . Hence the solution is . The second possibility is . This is easily solved for and the second solution is found from directly after that.

For the case when . 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 results in

(Remembering that now). Taking as the dependent variable and as the independent variable, then solving for gives

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

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

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

 number ode transformed type d’Alembert d’Alembert Clairaut d’Alembert Clairaut Clairaut d’Alembert d’Alembert

Notice in the above table, that the ODE is Clairaut when and d’Alembert otherwise.

#### 1.1 Example 1

To solve , it is first converted to which gives

 (1)

Using then (1) becomes

Hence and . Since then this is d’Alembert. Therefore equation (4) above is used to solve the ode. (4)  becomes

The solution to the above ODE is . Now is solved for in terms of (this is hard step in this algorithm). This might not always be possible and might have to be used. But in this example, the solution is

Substituting the above back into (2), since , gives the solution directly

#### 1.2 Example 2

To solve , it is first converted to which gives

Using then (1) becomes

Hence and . Since then this is d’Alembert. Therefore (4) becomes

This is linear in Solving the above gives

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

#### 1.3 Example 3

Solving . This is written as

Let then (1) becomes

Therefore and . Since then this is Clairaut ODE. This is special case. From (3),

First possibility is   which gives where and in this case looking at (1). Hence . Therefore the solution is

The other solution from considering or which implies . Substituting this into (2) gives 2 additional solutions

Therefore, the three solutions are

The solutions are singular solutions, since they can not be obtained from the general solution by giving a specific value for .

#### 1.4 Example 4

Solving . It is transformed to

Hence and . Since then this is d’Alembert. Therefore (4) becomes

This is linear in . Its solution is

Now is solved in terms of . This gives

 (2)

Hence the solution now found for from (1)

#### 1.5 Example 5

Solving . From above, it is transformed to

Therefore and . Since then this is Clairaut ODE. This is special case. From (3),

First possibility is   which gives where and in this case looking at (1). Hence . Therefore the solution is

The other solution from considering or , hence . Therefore from (1)

Therefore the solutions are

The solution is singular since it can not be obtained from .

#### 1.6 Example 6

Solving . From above, it is transformed to

Therefore and . Since then this is Clairaut ODE. This is special case. From (3),

First possibility is  which gives where and in this case looking at (1). Hence . Therefore the solution is

The other solution from considering or , hence . Therefore from (1)

Therefore the solutions are

The solution is singular since it can not be obtained from .

#### 1.7 Example 7

Solving . From above, it is transformed to

Hence and . Since then this is d’Alembert. Therefore (4) becomes

This is linear in . Solving for gives

Hence solving for gives

From (1) the solutions are

And

#### 1.8 Example 8

Solving . It is transformed to

Hence and . Since then this is d’Alembert. Therefore (4) becomes

Hence .  Solving for in terms of gives .  Substituting this in (1) gives the solution

#### 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.