# Differential Equations

## First Order Equations # Lagrange and Clairaut Equations

## Lagrange Equation

A differential equation of type

$y = x\varphi \left( {y'} \right) + \psi \left( {y'} \right),$

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:

$\left\{ \begin{array}{l} x = f\left( {p,C} \right)\\ y = f\left( {p,C} \right)\varphi \left( p \right) + \psi \left( p \right) \end{array} \right.$

provided that

$\varphi \left( p \right) - p \ne 0,$

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:

$y = \varphi \left( c \right)x + \psi \left( c \right),$

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

## Clairaut Equation

The Clairaut equation has the form:

$y = xy' + \psi \left( {y'} \right),$

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

$y = Cx + \psi \left( C \right),$

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:

$\left\{ \begin{array}{l} x = - \psi'\left( p \right)\\ y = xp + \psi \left( p \right) \end{array} \right.,$

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:

$y = 2xp - 3{p^2}.$

Differentiating both sides, we find:

$dy = 2xdp + 2pdx - 6pdp.$

We can replace $$dy$$ with $$pdx:$$

$pdx = 2xdp + 2pdx - 6pdp,\;\; \Rightarrow - pdx = 2xdp - 6pdp.$

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

$- dx = \frac{{2x}}{p}dp - 6dp,\;\; \Rightarrow \frac{{dx}}{{dp}} + \frac{2}{p}x - 6 = 0.$

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

$u\left( p \right) = \exp \left( {\int {\frac{2}{p}dp} } \right) = \exp \left( {2\ln \left| p \right|} \right) = \exp \left( {\ln {{\left| p \right|}^2}} \right) = {\left| p \right|^2} = {p^2}.$

The general solution of the linear equation is given by

$x\left( p \right) = \frac{{\int {{p^2} \cdot 6dp} + C}}{{{p^2}}} = \frac{{\frac{{6{p^3}}}{3} + C}}{{{p^2}}} = 2p + \frac{C}{{{p^2}}}.$

Substituting this expression for $$x$$ into the Lagrange equation, we obtain:

$y = 2\left( {2p + \frac{C}{{{p^2}}}} \right)p - 3{p^2} = 4{p^2} + \frac{{2C}}{p} - 3{p^2} = {p^2} + \frac{{2C}}{p}.$

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

$\left\{ \begin{array}{l} x\left( p \right) = 2p + \frac{C}{{{p^2}}}\\ y\left( p \right) = {p^2} + \frac{{2C}}{p} \end{array} \right..$

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

$2p - p = 0,\;\; \Rightarrow p = 0.$

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

$y = \varphi \left( 0 \right)x + \psi \left( 0 \right) = 0 \cdot x + 0 = 0.$

### 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:

$2y = 4xp + \ln p.$

Differentiate both sides of the equation:

$2dy = 4xdp + 4pdx + \frac{{dp}}{p}.$

As $$dy = pdx,$$ we get

$2pdx = 4xdp + 4pdx + \frac{{dp}}{p},\;\; \Rightarrow - 2pdx = 4xdp + \frac{{dp}}{p},\;\; \Rightarrow - 2p\frac{{dx}}{{dp}} = 4x + \frac{1}{p},\;\; \Rightarrow \frac{{dx}}{{dp}} + \frac{2}{p}x = - \frac{1}{{2{p^2}}}.$

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:

$u\left( p \right) = \exp \left( {\int {\frac{2}{p}dp} } \right) = \exp \left( {2\ln \left| p \right|} \right) = \exp \left( {\ln {{\left| p \right|}^2}} \right) = {\left| p \right|^2} = {p^2}.$

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

$x\left( p \right) = \frac{{\int {{p^2} \cdot \left( { - \frac{1}{{2{p^2}}}} \right)dp} + C}}{{{p^2}}} = \frac{{ - \frac{p}{2} + C}}{{{p^2}}} = - \frac{1}{{2p}} + \frac{C}{{{p^2}}}.$

Substituting this into the original equation, we find the parametric expression for $$y:$$

$2y = 4xp + \ln p,\;\; \Rightarrow 2y = 4p\left( { - \frac{1}{{2p}} + \frac{C}{{{p^2}}}} \right) + \ln p,\;\; \Rightarrow 2y = - 2 + \frac{{4C}}{p} + \ln p,\;\; \Rightarrow y = \frac{{2C}}{p} - 1 + \frac{{\ln p}}{2}.$

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

$\left\{ \begin{array}{l} x\left( p \right) = \frac{C}{{{p^2}}} - \frac{1}{{2p}}\\ y\left( p \right) = \frac{{2C}}{p} - 1 + \frac{{\ln p}}{2} \end{array} \right..$

To find the singular solution, we solve the equation:

$\varphi \left( p \right) - p = 0,\;\; \Rightarrow 2p - p = 0,\;\; \Rightarrow p = 0.$

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 $$p.$$

See more problems on Page 2.