# Differential Equations

## First Order Equations # 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
$u\left( {x,y} \right) = C.$

#### 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 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.

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Problem 1
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Problems 2-6