Section 1.6. Exact Differential Equations

Objective:

1. Partial derivative

2. Definition of exact differential equations

3. Solve the exact differential equation

We have learned how to solve first order linear ODE using integrating factor. When it is nonlinear but separable, then we can solve the ODE using separation of variables. At this section, we are solving ODE that is nonlinear and is not separable. First we need to learn how to take partial derivative. 

Definition: Let [latex]f(x,y)[/latex] be a function of two variables, [latex]x[/latex] and [latex]y[/latex]. The partial derivative function of [latex]f(x,y)[/latex] with respect to [latex]x[/latex] is 

\[f_{x}(x,y)=\lim_{h\rightarrow0}\frac{f(x+h,y)-f(x,y)}{h}=\frac{\partial f}{\partial x}\]

if the limit exists. This definition means that we take derivative of [latex]f(x,y)[/latex] along the path parallel to the [latex]x[/latex]-axis. We treat [latex]y[/latex]-variable as a constant and we differentiate [latex]f(x,y)[/latex] with respect to [latex]x[/latex]. The partial derivative function of [latex]f(x,y)[/latex] with respect to [latex]y[/latex] is 

\[f_{y}(x,y)=\lim_{h\rightarrow0}\frac{f(x,y+h)-f(x,y)}{h}=\frac{\partial f}{\partial y}\]

if the limit exists. This definition means that we take derivative of [latex]f(x,y)[/latex] along the path parallel to the [latex]y[/latex]-axis. We treat [latex]x[/latex]-variable as a constant and we differentiate [latex]f(x,y)[/latex] with respect to [latex]y[/latex]. 

Example 1: Find [latex]f_{x}(x,y)[/latex] and [latex]f_{y}(x,y)[/latex] where [latex]f(x,y)=x^{2}y-xy^{2}+xy+2x^{2}-y+2.[/latex]

Exercise 1: Find [latex]f_{x}(x,y)[/latex] and [latex]f_{y}(x,y)[/latex] where [latex]f(x,y)=2x^{2}y+xy^{2}-3xy-2x+y^{2}-3.[/latex] }

Example 2: Find [latex]f_{x}(x,y)[/latex], and [latex]f_{y}(x,y)[/latex] where [latex]f(x,y)=e^{x^{2}y}+\text{sin}(xy^{2}).[/latex]

Exercise 2: Find [latex]f_{x}(x,y)[/latex], and [latex]f_{y}(x,y)[/latex] where [latex]f(x,y,z)=e^{xy}+\text{ln}(xy).[/latex]}

Theorem: Suppose an ODE can be written in the form [latex]M(x,y)+N(x,y)y'=0[/latex] where the functions [latex]M,N,[/latex] [latex]M_{y}[/latex] and [latex]N_{x}[/latex] are all continuous in the rectangular region [latex]R[/latex]: [latex]a\lt x \lt b[/latex],  [latex]c\lt y \lt d[/latex]. Then the equation is an exact differential equation if and only if [latex]M_{y}(x,y)=N_{x}(x,y)[/latex], for all [latex](x,y)[/latex] in [latex]\mathbb{R}^{2}[/latex] if and only if there exists a function [latex]\phi[/latex] satisfying the conditions [latex]\phi_{x}(x,y)=M(x,y)[/latex] and [latex]\phi_{y}(x,y)=N(x,y)[/latex].

Example 3: [latex](4x-1)+(y+2)y'=0[/latex]. Show the ODE is exact and find its general solution. 

Exercise 3: [latex](2x+5)+(3y^{2}+2y)y'=0[/latex]. Show the ODE is exact and find its general solution. 

Example 4: [latex](4x-y^{2})+(-2xy+2y)y'=0[/latex]. Show the ODE is exact and find its general solution. 

Exercise 4: [latex](3x^{2}+y)+(x+3y^{2})y'=0[/latex]. Show the ODE is exact and find its general solution. 

Example 5: [latex](\text{sin}(x)+e^{y})+(xe^{y}+3y^{2})y'=0[/latex].

[latex]y(0)=0[/latex]. Show the ODE is exact and find the solution of IVP. 

Exercise 5: [latex](e^{x}+3y^{2})+(6xy+3e^{y})y'=0[/latex], [latex]y(0)=1[/latex]. Show the ODE is exact and find the solution of IVP.

Group Work

1. Find [latex]b[/latex] such that the ODE is exact then find the general solution. [latex](xy^{2}+bx^{2}y)+(x+y)x^{2}y'=0[/latex].

2. Show the ODE is exact and Solve the IVP.  [latex](4x^{3}-y^{2}+2xy+1)+(-2xy+x^{2}-1+3y^{2})y'=0[/latex], [latex]y(1)=0[/latex].

3. Show the ODE is exact and find its general solution. [latex](3y\text{cos}(x)+4xe^{x}+2x^{2}e^{x})dx+(3\text{sin}(x)+3)dy=0[/latex].

4. Show the ODE is exact and Solve the IVP. [latex](-4y\text{cos}(x)+4\text{sin}(x)\text{cos}(x)+\text{sec}^{2}(x))dx+(4y-4\text{sin}(x))dy=0,y(\frac{\pi}{3})=0[/latex].

5. Show the ODE is exact and Solve the IVP. [latex](y^{3}-1)e^{x}dx+3y^{2}(e^{x}+1)dy=0,y(0)=0.[/latex]