Differential Equations

First Order Equations

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Lagrange and Clairaut Equations

Lagrange Equation

A differential equation of type

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

where φ (y' ) and ψ (y' ) 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 \[y = 2xy' - 3{\left( {y'} \right)^2}.\]


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 \[2y - 4xy' - \ln y' = 0.\]


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

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

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 check for the possible singular solution solve the equation:

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

This yields:

\[y = \varphi \left( 0 \right)x + \psi \left( 0 \right) = 0 \cdot x + \frac{1}{2}\ln 0 \to - \infty .\]

Thus, the differential equation has no singular solution.

See more problems on Page 2.

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