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:

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