Differential Equations

First Order Equations

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Exact Differential Equations

  • Definition of Exact Equation

    A differential equation of type

    \[{P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy }={ 0}\]

    is called an exact differential equation if there exists a function of two variables \(u\left( {x,y} \right)\) with continuous partial derivatives such that

    \[{du\left( {x,y} \right) \text{ = }}\kern0pt{ P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy.}\]

    The general solution of an exact equation is given by

    \[u\left( {x,y} \right) = C,\]

    where \(C\) is an arbitrary constant.

    Test for Exactness

    Let functions \(P\left( {x,y} \right)\) and \(Q\left( {x,y} \right)\) have continuous partial derivatives in a certain domain \(D.\) The differential equation \(P\left( {x,y} \right)dx +\) \( Q\left( {x,y} \right)dy \) \(= 0\) is an exact equation if and only if

    \[\frac{{\partial Q}}{{\partial x}} = \frac{{\partial P}}{{\partial y}}.\]

    Algorithm for Solving an Exact Differential Equation

    1. First it’s necessary to make sure that the differential equation is exact using the test for exactness:
      \[\frac{{\partial Q}}{{\partial x}} = \frac{{\partial P}}{{\partial y}}.\]
    2. Then we write the system of two differential equations that define the function \(u\left( {x,y} \right):\)
      \[\left\{ \begin{array}{l} \frac{{\partial u}}{{\partial x}} = P\left( {x,y} \right)\\ \frac{{\partial u}}{{\partial y}} = Q\left( {x,y} \right) \end{array} \right..\]
    3. Integrate the first equation over the variable \(x.\) Instead of the constant \(C,\) we write an unknown function of \(y:\)
      \[{u\left( {x,y} \right) \text{ = }}\kern0pt{ \int {P\left( {x,y} \right)dx} + \varphi \left( y \right).}\]
    4. Differentiating with respect to \(y,\) we substitute the function \(u\left( {x,y} \right)\)into the second equation:
      \[ {\frac{{\partial u}}{{\partial y}} \text{ = }}\kern0pt {\frac{\partial }{{\partial y}}\left[ {\int {P\left( {x,y} \right)dx} + \varphi \left( y \right)} \right] } = {Q\left( {x,y} \right).} \]
      From here we get expression for the derivative of the unknown function \({\varphi \left( y \right)}:\)
      \[ {\varphi’\left( y \right) } = {Q\left( {x,y} \right) }-{ \frac{\partial }{{\partial y}}\left( {\int {P\left( {x,y} \right)dx} } \right).} \]
    5. By integrating the last expression, we find the function \({\varphi \left( y \right)}\) and, hence, the function \(u\left( {x,y} \right):\)
      \[{u\left( {x,y} \right) \text{ = }}\kern0pt{ \int {P\left( {x,y} \right)dx} + \varphi \left( y \right).}\]
    6. The general solution of the exact differential equation is given by
      \[u\left( {x,y} \right) = C.\]


    In Step \(3,\) we can integrate the second equation over the variable \(y\) instead of integrating the first equation over \(x.\) After integration we need to find the unknown function \({\psi \left( x \right)}.\)

  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Solve the differential equation \(2xydx +\) \( \left( {{x^2} + 3{y^2}} \right)dy \) \(= 0.\)

    Example 2

    Find the solution of the differential equation \(\left( {6{x^2} – y + 3} \right)dx +\) \( \left( {3{y^2} – x – 2} \right)dy \) \(= 0.\)

    Example 3

    Solve the differential equation \({e^y}dx +\) \(\left( {2y + x{e^y}} \right)dy \) \(= 0.\)

    Example 4

    Solve the equation \(\left( {2xy – \sin x} \right)dx +\) \( \left( {{x^2} – \cos y} \right)dy \) \(= 0.\)

    Example 5

    Solve the equation \(\left( {1 + 2x\sqrt {{x^2} – {y^2}} } \right)dx -\) \( 2y\sqrt {{x^2} – {y^2}} dy \) \(= 0.\)

    Example 6

    Solve the differential equation \({\large\frac{1}{{{y^2}}}\normalsize} – {\large\frac{2}{x}\normalsize} =\) \( {\large\frac{{2xy’}}{{{y^3}}}\normalsize}\) with the initial condition \(y\left( 1 \right) = 1.\)

    Example 1.

    Solve the differential equation \(2xydx +\) \( \left( {{x^2} + 3{y^2}} \right)dy \) \(= 0.\)


    The given equation is exact because the partial derivatives are the same:

    {{\frac{{\partial Q}}{{\partial x}} }={ \frac{\partial }{{\partial x}}\left( {{x^2} + 3{y^2}} \right) }={ 2x,\;\;}}\kern-0.3pt
    {{\frac{{\partial P}}{{\partial y}} }={ \frac{\partial }{{\partial y}}\left( {2xy} \right) }={ 2x.}}

    We have the following system of differential equations to find the function \(u\left( {x,y} \right):\)

    \[\left\{ \begin{array}{l} \frac{{\partial u}}{{\partial x}} = 2xy\\ \frac{{\partial u}}{{\partial y}} = {x^2} + 3{y^2} \end{array} \right..\]

    By integrating the first equation with respect to \(x,\) we obtain

    \[{u\left( {x,y} \right) = \int {2xydx} }={ {x^2}y + \varphi \left( y \right).}\]

    Substituting this expression for \(u\left( {x,y} \right)\) into the second equation gives us:

    {{\frac{{\partial u}}{{\partial y}} }={ \frac{\partial }{{\partial y}}\left[ {{x^2}y + \varphi \left( y \right)} \right] }={ {x^2} + 3{y^2},\;\;}}\Rightarrow
    {{{x^2} + \varphi’\left( y \right) }={ {x^2} + 3{y^2},\;\;}}\Rightarrow
    {\varphi’\left( y \right) = 3{y^2}.}

    By integrating the last equation, we find the unknown function \({\varphi \left( y \right)}:\)

    \[\varphi \left( y \right) = \int {3{y^2}dy} = {y^3},\]

    so that the general solution of the exact differential equation is given by

    \[{x^2}y + {y^3} = C,\]

    where \(C\) is an arbitrary constant.

    Page 1
    Problem 1
    Page 2
    Problems 2-6