Exact Equations In Calculus

Lecture 1.6: Exact Equations in Calculus

Introduction to Exact Equations

• Exact differential equations are a specific type of differential equation that can be solved under certain conditions involving differentiable functions.

• A multivariable function is defined such that if $f: ext{R}^2 \to \text{R}$, then we can express this relationship as:

  • $z = f(x, y)$

• The associated differential is given by:

  • $dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$

Level Curves

• A level curve of a function $f(x, y)$ is defined as a set of points where the function takes on a constant value.

  • If $C$ is a constant, then the level curves are represented by:

  • $f(x, y) = C$

• An example is:

  • For $f(x, y) = x^2 + y^2$, level curves are expressed as:

  • $f(x, y) = 1$ (circle with radius 1)

  • $f(x, y) = 3$ (circle with radius $\sqrt{3}$)

Differential Equations and Exactness

• A differential equation of the form:

  • $M(x, y) dx + N(x, y) dy = 0$

• is called exact if there exists a function $f(x, y)$ such that:

  • $M(x, y) = \frac{\partial f}{\partial x}$

  • $N(x, y) = \frac{\partial f}{\partial y}$

• If this is satisfied, then the equation can be expressed in terms of a constant function:

  • $f(x, y) = C$

Identification of Exact Equations

1. Given:

   • $M(x,y) dx + N(x,y) dy = 0$

2. Check if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.

3. If they are equal, the equation is exact.

   • Example: For $2x dx + 2y dy = 0$:

   • $M(x, y) = 2x$

   • $N(x, y) = 2y$

4. Differentiate:

   • $\frac{\partial M}{\partial y} = 0$

   • $\frac{\partial N}{\partial x} = 0$

5. Conclusion: Since they are equal, this confirms that the equation is exact.

Examples of Exact Equations

Example 1:

• Consider the equation:

  • $x^2 y^3 dx + x^3 y^2 dy = 0$

• Here,

  • $M(x,y) = x^2y^3$

  • $N(x,y) = x^3y^2$

• Check for exactness:

  • $\frac{\partial M}{\partial y} = 3x^2y^2$

  • $\frac{\partial N}{\partial x} = 2x^2y^3$

• Since they are equal, this differential equation is exact.

Example 2:

• Differential:

  • $2xy dx + (x^2 - 1) dy = 0$

• Where,

  • $M(x,y) = 2xy$

  • $N(x,y) = x^2 - 1$

• Check for exactness:

  • $\frac{\partial M}{\partial y} = 2x$

  • $\frac{\partial N}{\partial x} = 2x$

Solving Exact Differential Equations

1. Identify $M(x,y)$ and $N(x,y)$ from the equation:

   • Given $M(x,y) dx + N(x,y) dy = 0$

2. Compute:

   • $\frac{\partial f}{\partial x} = M(x,y)$

   • $\frac{\partial f}{\partial y} = N(x,y)$

3. Integrate $M(x,y)$ with respect to $x$ and add an arbitrary function of $y$:

   • $f(x,y) = \int M(x,y) dx + g(y)$

4. Differentiate $f(x,y)$ with respect to $y$ to find $g'(y)$.

5. Set it equal to $N(x,y)$ and solve for $g(y)$.

Example:

• Solve the exact DE:

  • $2xy dx + (x^2 - 1) dy = 0$

  1. Identify:

  • $M(x,y) = 2xy$

  • $N(x,y) = x^2 - 1$

  1. Integrate for $f(x,y)$:

  • $f(x,y) = x^2y + g(y)$

  1. Find $g'(y)$ and solve for $g(y)$.

Closing Remarks

• A 5-step process for solving exact equations has been established:

  1. Identify $M(x,y)$ and $N(x,y)$.

  2. Check for exactness.

  3. Integrate to find $f(x,y)$.

  4. Differentiate and determine the function $g(y)$.

  5. Solve the implicit equation $f(x,y) = C$ for the solution.