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 \(\varphi \left( {y’} \right)\) 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 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 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
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    Problems 2-4