First Order Differential Equations
2.3 Exact Differential Equations
A. Introduction
Exact differential equations are a class of first-order differential equations that can be solved using a particular integrability condition. This section will discuss what makes an equation exact, how to verify this condition, and the methodology for solving such equations.
We begin by introducing a foundational theorem followed by an illustrative example to demonstrate its application. Following that, we delve into the concept of exact equations and explore a method for solving them.
Theorem: If function [asciimath]F(x,y)[/asciimath] has continuous partial derivatives [asciimath]F_x[/asciimath] and [asciimath]F_y[/asciimath], then the equation [asciimath]F(x,y)=c[/asciimath] is an implicit solution to the differential equation [asciimath]F_x(x,y) dx + F_y(x,y) dy = 0[/asciimath].
The theorem can be proven by using implicit differentiation.
Show that [asciimath]x^2 y^3+xy^3+3xy=c[/asciimath] is an implicit solution for the given differential equation.
[asciimath](2x y^3+y^3+3y)dx+(3x^2 y^2+3xy^2+3x)dy=0[/asciimath]
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To apply the theorem effectively, we need to define [asciimath]F(x,y)[/asciimath] as the function given in the solution. Then, we show that the terms multiplied by dx and dy are, respectively, the partial derivatives [asciimath]F_x[/asciimath] and [asciimath]F_y[/asciimath] of [asciimath]F[/asciimath] with respect to [asciimath]x[/asciimath] and [asciimath]y[/asciimath]. This process involves finding these partial derivatives and confirming that they correspond to the respective terms in the given differential equation.
letting [asciimath]F(x,y)=[/asciimath] [asciimath]x^2 y^3+xy^3+3xy[/asciimath] , we find its partial derivatives:
[asciimath]F_x=2xy^3+y^3+3y[/asciimath]
[asciimath]F_y=3x^2y^2+3xy^2+3x[/asciimath]
We observe that [asciimath]F_x[/asciimath] and [asciimath]F_y[/asciimath] are equivalent to the expressions multiplied by [asciimath]dx[/asciimath] and [asciimath]dy[/asciimath] in the equation, respectively, which confirms that [asciimath]F(x,y)=c[/asciimath] is the solution to the given differential equation.
[asciimath]underbrace((2x y^3+y^3+3y))_(F_x)dx+underbrace((3x^2 y^2+3xy^2+3x))_(F_y)dy=0[/asciimath]
B. Solution to Exact Equations
We now shift our focus to a broader understanding of exact differential equations. Consider a differential equation expressed as
[asciimath]M(x,y)dx + N(x,y) dy = 0[/asciimath]
which can also be represented as
[asciimath]M(x,y) + N(x,y) (dy)/(dx) = 0[/asciimath].
An equation of this form is called exact if there is a function [asciimath]F(x,y)[/asciimath] such that its partial derivatives [asciimath]F_x[/asciimath] and [asciimath]F_y[/asciimath] correspond to [asciimath]M(x,y)[/asciimath] and [asciimath]N(x,y)[/asciimath], respectively. When such a function exists[asciimath], F(x,y)=c[/asciimath] represents a solution to the differential equation.
For instance, the equations [asciimath]4xy^2dx-7x^3ydy=0[/asciimath] and [asciimath]5ysinx-xycosx(dy)/(dx)=0[/asciimath] are examples of equations in exact form.
Now the pertinent questions are
- How can we determine whether a given differential equation is exact?
- If it is exact, how do we find the function [asciimath]F(x,y)[/asciimath] and thus a solution?
[asciimath]F_(xy)=[/asciimath] [asciimath]F_(yx)[/asciimath]
or equivalently,
[asciimath]M_y=N_x[/asciimath]
This relationship is summarized in the theorem below.
1) Test for Exactness
Theorem. Consider that the first derivatives of [asciimath]M (x,y)[/asciimath] and [asciimath]N (x,y)[/asciimath] are continuous within a rectangular region [asciimath]RR[/asciimath]. Then, the differential equation
[asciimath]M(x,y)dx+N(x,y)dy=0[/asciimath]
is exact in [asciimath]RR[/asciimath] if, and only if, the following condition is satisfied for all [asciimath](x,y)[/asciimath] in [asciimath]RR[/asciimath]:
[asciimath](delM)/(dely) (x,y)=(delN)/(delx) (x,y)[/asciimath]
To address the second question of solving an exact differential equation, follow the step-by-step procedure outlined below.
2) Method for Solving Exact Equations
1*. Find [asciimath]F(x,y)[/asciimath]: If the equation is exact, then [asciimath](delF)/(delx) (x,y)=M[/asciimath]. Integrate this equation with respect to [asciimath]x[/asciimath] to find part of [asciimath]F[/asciimath]. Remember to include an arbitrary function of the other variable, in this case [asciimath]y[/asciimath].
[asciimath]F(x,y)=int M(x,y)dx+g(y)[/asciimath]
2. Determine the Arbitrary Function:
a. To find [asciimath]g(y)[/asciimath], first determine [asciimath]F_y[/asciimath] from the expression obtained for F(x,y) in Step 1. Since [asciimath]F_y[/asciimath] must be equal to [asciimath]N(x,y)[/asciimath] from the exact differential equation, set [asciimath]F_y[/asciimath] equal to [asciimath]N(x,y)[/asciimath] and solve for [asciimath]g'(y)[/asciimath].
b. After isolating [asciimath]g'(y)[/asciimath], integrate it with respect to [asciimath]y[/asciimath] to obtain [asciimath]g(y)[/asciimath]. Set the constant of integration to zero. Substitute the determined [asciimath]g(y)[/asciimath] back into the expression for [asciimath]F(x,y)[/asciimath] to complete it.
3. Form the general Solution: The solution to [asciimath]M(x,y)dx + N(x,y) dy = 0[/asciimath] is given implicitly (not solved for [asciimath]y[/asciimath]) by
[asciimath]F(x,y)=C[/asciimath]
where [asciimath]C[/asciimath] is a constant. This equation represents the family of curves that are solutions to the differential equation.
*Note: As an alternative method, you might also start by integrating [asciimath](delF)/(dely) (x,y)=N[/asciimath] with respect to [asciimath]y[/asciimath] and then use similar steps to find [asciimath]F(x,y)[/asciimath] if the integration seems to be easier.
Determine if the equation is exact and if so find the solution: [asciimath]3y^3 dx + 9xy^2 dy = 0[/asciimath]
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1) Test for Exactness:
[asciimath]M=3y^3[/asciimath] [asciimath]->[/asciimath] [asciimath](delM)/(dely) (x,y)=9y^2[/asciimath]
[asciimath]N=9xy^2[/asciimath] [asciimath]->[/asciimath] [asciimath](delN)/(delx) (x,y) = 9y^2[/asciimath]
Since [asciimath]M_y=N_x[/asciimath], the equation is exact.
2) Find the solution:
1. We know [asciimath]F_x = M=3y^3[/asciimath] . We integrate with respect to [asciimath]x[/asciimath]:
[asciimath]F(x,y)=int3y^3dx+g(y)[/asciimath]
[asciimath]= 3xy^3 + g(y)[/asciimath]
2a. To find [asciimath]g(y)[/asciimath], we take the partial derivative of above [asciimath]F[/asciimath] with respect to [asciimath]y[/asciimath]:
[asciimath]F_y = 9xy^2 + g'(y)[/asciimath]
Since [asciimath]F_y[/asciimath] must be equal to [asciimath]N(x,y)=9xy^2[/asciimath] from the exact differential equation, set [asciimath]F_y[/asciimath] equal to [asciimath]N(x,y)[/asciimath] and solve for [asciimath]g'(y)[/asciimath]or determine it by comparing.
By comparing, we determine that [asciimath]g'(y) = 0[/asciimath].
2b. By integrating [asciimath]g'(y)[/asciimath] with respect to y, we obtain [asciimath]g(y)= C[/asciimath]. Setting the constant of integration to zero gives [asciimath]g(y)=0[/asciimath], resulting in [asciimath]F = 3xy^3[/asciimath].
3. Thus, an implicit solution to the differential equation is
[asciimath]3xy^3 = C[/asciimath]
Try an Example
Try an Example
a) Solve the initial value problem and find the explicit solution [asciimath]y=f(x)[/asciimath]. b) Determine the interval of validity.
[asciimath](3y^3-1)e^x dx+9y^2(e^x-3)dy = 0,[/asciimath] [asciimath]y(0)=2[/asciimath]
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a)
1) Test for Exactness:
[asciimath]M=(3y^3-1)e^x[/asciimath] [asciimath]->[/asciimath] [asciimath](delM)/(dely) (x,y)=9y^2e^x[/asciimath]
[asciimath]N=9y^2(e^x-3)[/asciimath] [asciimath]->[/asciimath] [asciimath](delN)/(delx) (x,y) = 9y^2e^x[/asciimath]
Since [asciimath]M_y=N_x[/asciimath], the equation is exact.
2) Find the general solution:
We have the option to integrate [asciimath]M[/asciimath] with respect to [asciimath]x[/asciimath] or integrate [asciimath]N[/asciimath] with respect to [asciimath]y[/asciimath]. Since both integrals are equally straightforward in this case, we integrate [asciimath]N[/asciimath] with respect to [asciimath]y[/asciimath] for variety, ensuring we provide examples of both methods.
1.
[asciimath](delF)/(dely) = N[/asciimath]
[asciimath]F(x,y)=int9y^2(e^x-3) dy+h(x)[/asciimath]
[asciimath]= 3y^3(e^x-3) + h(x)[/asciimath]
It is important to note that we include an arbitrary function of [asciimath]x[/asciimath], [asciimath]h(x)[/asciimath], since we integrate with respect to [asciimath]y[/asciimath] this time.
2a. To find [asciimath]h(x)[/asciimath], we take the partial derivative of above [asciimath]F[/asciimath] with respect to [asciimath]x[/asciimath]:
[asciimath]F_x = 3y^3e^x + h'(x)[/asciimath]
Since [asciimath]F_x[/asciimath] must be equal to [asciimath]M(x,y)=(3y^3-1)e^x[/asciimath] from the exact differential equation, we set [asciimath]F_x[/asciimath] equal to [asciimath]M(x,y)[/asciimath] and solve for [asciimath]h'(x)[/asciimath] or determine it by comparing.
[asciimath]F_x=M(x,y)[/asciimath]
[asciimath]3y^3e^x + h'(x)=[/asciimath] [asciimath](3y^3-1)e^x[/asciimath]
[asciimath]h'(x)=-e^x[/asciimath]
2b. By integrating [asciimath]h'(x)[/asciimath] with respect to [asciimath]x[/asciimath], we obtain [asciimath]h(x)= -e^x+C_1[/asciimath]. Setting the constant of integration to zero gives [asciimath]h(x)=-e^x[/asciimath]. Therefore,
[asciimath]F(x,y)= 3y^3(e^x-3) -e^x[/asciimath]
3. Thus, an implicit solution to the differential equation is
[asciimath]3y^3(e^x-3) -e^x = C[/asciimath]
Apply the initial condition:
[asciimath]y(0)=2[/asciimath]
[asciimath]3(2^3)(e^0-3) -e^0 = C[/asciimath]
[asciimath]24(1-3)-1=C[/asciimath]
[asciimath]C=-49[/asciimath]
The solution to the IVP problem then is
[asciimath]3y^3(e^x-3) -e^x = -49[/asciimath]
We need to find the explicit solution, so we rearrange the equation to solve for [asciimath]y[/asciimath]:
[asciimath]3y^3(e^x-3) = e^x-49[/asciimath]
[asciimath]y^3= (e^x-49)/(3(e^x-3))[/asciimath]
[asciimath]y=root(3)((e^x-49)/(3(e^x-3)) )[/asciimath]
b) Find the interval of validity:
To establish the interval of validity for the solution, we need to ensure the denominator of the rational function is not equal to zero to avoid undefined expressions:
[asciimath]e^x-3!=0[/asciimath]
[asciimath]x!=ln(3)[/asciimath]
Therefore, the interval of validity for the solution is
Try an Example
Section 2.3 Exercises
- Determine if the equation is exact and if so find the solution: [asciimath]-x y^2-4x y+(-x^2y-2x^2+3)(dy)/(dx) = 0[/asciimath].
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[asciimath]-1/2x^2y^2-2x^2y+3y=C[/asciimath]
- Solve the differential equation: [asciimath](dy)/(dx) = (-10 e^xcos(y)-3y^2/x)/(-10 e^xsin(y)+6yln(x)+2y^2)[/asciimath].
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[asciimath]10 e^xcos(y)+3y^2ln(x)+2/3y^3=C[/asciimath]
- Solve the initial value problem. Give the explicit solution: [asciimath](2y^3-1)e^x dx+6y^2(e^x+3)dy = 0[/asciimath], [asciimath]y(0)=-2[/asciimath].
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[asciimath]y = ((e^x-65)/(2 e^x+6))^(1/3)[/asciimath]
