Definition of Exact Equation
A differential equation of type
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[{Pleft( {x,y} right)dx + Qleft( {x,y} right)dy }={ 0}]
is called an exact differential equation if there exists a function of two variables (uleft( {x,y} right)) with continuous partial derivatives such that
[{duleft( {x,y} right) text{ = }}kern0pt{ Pleft( {x,y} right)dx + Qleft( {x,y} right)dy.}]
The general solution of an exact equation is given by
[uleft( {x,y} right) = C,]
where (C) is an arbitrary constant.
Test for Exactness
Let functions (Pleft( {x,y} right)) and (Qleft( {x,y} right)) have continuous partial derivatives in a certain domain (D.) The differential equation (Pleft( {x,y} right)dx +) ( Qleft( {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
- 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}}.]
- Then we write the system of two differential equations that define the function (uleft( {x,y} right):)[left{ begin{array}{l}frac{{partial u}}{{partial x}} = Pleft( {x,y} right)frac{{partial u}}{{partial y}} = Qleft( {x,y} right)end{array} right..]
- Integrate the first equation over the variable (x.) Instead of the constant (C,) we write an unknown function of (y:)[{uleft( {x,y} right) text{ = }}kern0pt{ int {Pleft( {x,y} right)dx} + varphi left( y right).}]
- Differentiating with respect to (y,) we substitute the function (uleft( {x,y} right))into the second equation:[{frac{{partial u}}{{partial y}} text{ = }}kern0pt{frac{partial }{{partial y}}left[ {int {Pleft( {x,y} right)dx} + varphi left( y right)} right] }= {Qleft( {x,y} right).}]From here we get expression for the derivative of the unknown function ({varphi left( y right)}:)[{varphi’left( y right) }= {Qleft( {x,y} right) }-{ frac{partial }{{partial y}}left( {int {Pleft( {x,y} right)dx} } right).}]
- By integrating the last expression, we find the function ({varphi left( y right)}) and, hence, the function (uleft( {x,y} right):)[{uleft( {x,y} right) text{ = }}kern0pt{ int {Pleft( {x,y} right)dx} + varphi left( y right).}]
- The general solution of the exact differential equation is given by
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
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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 + 2xsqrt {{x^2} – {y^2}} } right)dx -) ( 2ysqrt {{x^2} – {y^2}} dy ) (= 0.)Example 6
Solve the differential equation ({largefrac{1}{{{y^2}}}normalsize} – {largefrac{2}{x}normalsize} =) ( {largefrac{{2xy’}}{{{y^3}}}normalsize}) with the initial condition (yleft( 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.}}
]
{{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 (uleft( {x,y} right):)
[left{ begin{array}{l}frac{{partial u}}{{partial x}} = 2xyfrac{{partial u}}{{partial y}} = {x^2} + 3{y^2}end{array} right..]
By integrating the first equation with respect to (x,) we obtain
[{uleft( {x,y} right) = int {2xydx} }={ {x^2}y + varphi left( y right).}]
Substituting this expression for (uleft( {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}.}
]
{{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.
Problem 1
Problems 2-6
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