# Lagrange and Clairaut Equations

## Lagrange Equation

A differential equation of type

where \(\varphi \left( {y'} \right)\) and \(\psi \left( {y'} \right)\) are known functions differentiable on a certain interval, is called the Lagrange equation.

By setting \(y' = p\) and differentiating with respect to \(x,\) we get the general solution of the equation in parametric form:

provided that

where \(p\) is a parameter.

Lagrange equation may also have a singular solution if the condition \(\varphi \left( p \right) - p\) \( \ne 0\) is failed. The singular solution is given by the expression:

where \(c\) is the root of the equation \(\varphi \left( p \right) - p = 0.\)

## Clairaut Equation

The Clairaut equation has the form:

here \(\psi \left( {y'} \right)\) is a nonlinear differentiable function. The Clairaut equation is a particular case of the Lagrange equation when \(\varphi \left( {y'} \right) = y'.\) It is solved in the same way by introducing a parameter. The general solution is given by

where \(C\) is an arbitrary constant.

Similarly to the Lagrange equation, the Clairaut equation may have a singular solution that is expressed parametrically in the form:

where \(p\) is a parameter.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the general and singular solutions of the differential equation \[y = 2xy' - 3{\left( {y'} \right)^2}.\]

### Example 2

Find the general and singular solutions of the equation \[2y - 4xy' - \ln y' = 0.\]

### Example 1.

Find the general and singular solutions of the differential equation \[y = 2xy' - 3{\left( {y'} \right)^2}.\]

Solution.

Here we see that we deal with a Lagrange equation. We will solve it using the method of differentiation.

Denote \(y' = p,\) so the equation is written in the form:

Differentiating both sides, we find:

We can replace \(dy\) with \(pdx:\)

By dividing by \(p,\) we can write the following equation (later we check if \(p = 0\) is a solution of the original equation):

As it can be seen, we obtain a linear equation for the function \(x\left( p \right).\) The integrating factor is

The general solution of the linear equation is given by

Substituting this expression for \(x\) into the Lagrange equation, we obtain:

Thus, the general solution in parametric form is defined by the system of equations:

Besides that, the Lagrange equation can have a singular solution. Solving the equation \(\varphi \left( p \right) - p = 0,\) we find the root:

Hence, the singular solution is expressed by the linear function:

### Example 2.

Find the general and singular solutions of the equation \[2y - 4xy' - \ln y' = 0.\]

Solution.

Here we have a Lagrange equation. By setting \(y' = p,\) we can write:

Differentiate both sides of the equation:

As \(dy = pdx,\) we get

When we divided by \(p\), we lost the root \(p = 0,\) which corresponds to the solution \(y = 0.\)

Thus, we get a linear differential equation for the function \(x\left( p \right).\) We solve it using the integrating factor:

The function \(x\left( p \right)\) is defined by

Substituting this into the original equation, we find the parametric expression for \(y:\)

Hence, the general solution in parametric form is written as follows:

To find the singular solution, we solve the equation:

It follows from this that \(y = C.\) We can make direct substitution to make sure that the constant \(C\) is equal to zero.

Thus, the differential equation has the singular solution \(y = 0.\) We have already met with this solution above when we divided the equation by \(p.\)