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# Differential Equations

1st Order Equations

# Lagrange and Clairaut Equations

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Problem 1
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Problems 2-4

### 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:

$y = xy’ + \psi \left( {y’} \right),$

where $$\psi \left( {y’} \right)$$ 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 on problem description to see 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 3

Find the general and singular solutions of the differential equation $$y = xy’ + {\left( {y’} \right)^2}.$$

### ✓Example 4

Find the general and singular solutions of the differential equation $$y = xy’ +$$ $$\sqrt {{{\left( {y’} \right)}^2} + 1} .$$

### 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:

${y = \varphi \left( 0 \right)x + \psi \left( 0 \right) }={ 0 \cdot x + 0 }={ 0.}$
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Problem 1
Page 2
Problems 2-4