Differential Equations

1st Order Equations

Exact Differential Equations

Page 1
Problem 1
Page 2
Problems 2-6

Definition of Exact Equation

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

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 on problem description to see 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 \(2xydx +\) \( \left( {{x^2} + 3{y^2}} \right)dy \) \(= 0.\)

Solution.

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

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

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

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