Differential Equations

First Order Equations

1st Order Diff Equations Logo

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

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

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 \(y:\)
    \[u\left( {x,y} \right) = \int {P\left( {x,y} \right)dx} + \varphi \left( y \right).\]
  4. Differentiating with respect to \(y,\) we substitute the function \(u\left( {x,y} \right)\)into the second equation:
    \[\frac{{\partial u}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\int {P\left( {x,y} \right)dx} + \varphi \left( y \right)} \right] = Q\left( {x,y} \right).\]
    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 \(u\left( {x,y} \right):\)
    \[u\left( {x,y} \right) = \int {P\left( {x,y} \right)dx} + \varphi \left( y \right).\]
  6. The general solution of the exact differential equation is given by
    \[u\left( {x,y} \right) = C.\]

Note:

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 or tap 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 1.

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

Solution.

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

\[\frac{{\partial Q}}{{\partial x}}= \frac{\partial }{{\partial x}}\left( {{x^2} + 3{y^2}} \right) = 2x,\;\; \frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {2xy} \right) = 2x.\]

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.

Example 2.

Find the solution of the differential equation

\[\left( {6{x^2} - y + 3} \right)dx + \left( {3{y^2} - x - 2} \right)dy = 0.\]

Solution.

We check this equation for exactness:

\[\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {3{y^2} - x - 2} \right) = - 1,\;\; \frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {6{x^2} - y + 3} \right) = - 1.\]

Hence, the given differential equation is exact. Write the system of equations to determine the function \(u\left( {x,y} \right):\)

\[\left\{ \begin{array}{l} \frac{{\partial u}}{{\partial x}} = P\left( {x,y} \right) = 6{x^2} - y + 3\\ \frac{{\partial u}}{{\partial y}} = Q\left( {x,y} \right) = 3{y^2} - x - 2 \end{array} \right.\]

Integrate the first equation with respect to the variable \(x\) assuming that \(y\) is a constant. This produces:

\[u\left( {x,y} \right) = \int {\left( {6{x^2} - y + 3} \right)dx} = \frac{{6{x^3}}}{3} - xy + 3x + \varphi \left( y \right) = 2{x^3} - xy + 3x + \varphi \left( y \right).\]

Here we introduced a continuous differentiable function \(\varphi \left( y \right)\) instead of the constant \(C.\)

Plug in the function \(u\left( {x,y} \right)\) into the second equation:

\[\frac{{\partial u}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {2{x^3} - xy + 3x + \varphi \left( y \right)} \right] = - \cancel{x} + \varphi'\left( y \right) = 3{y^2} - \cancel{x} - 2.\]

We get equation for the derivative \(\varphi'\left( y \right):\)

\[\varphi'\left( y \right) = 3{y^2} - 2.\]

Integrating gives the function \(\varphi \left( y \right):\)

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

So, the function \(u\left( {x,y} \right)\) is given by

\[u\left( {x,y} \right) = 2{x^3} - xy + 3x + {y^3} - 2y.\]

Hence, the general solution of the equation is defined by the following implicit expression:

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

where \(C\) is an arbitrary real number.

See more problems on Page 2.

Page 1 Page 2