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Lesson 3: Exact Equations

Estimated time: 35-40 minutes

Learning Objectives

Write a first-order ODE in the differential form M(x,y) dx + N(x,y) dy = 0
Apply the exactness test: ∂M/∂y = ∂N/∂x
Solve exact equations by finding the potential function F(x,y)
Recognize when an equation is not exact

The Idea Behind Exact Equations

If there exists a function F(x,y) such that dF = M dx + N dy, then the equation M dx + N dy = 0 is simply dF = 0, whose solution is F(x,y) = C. Such an equation is called exact.

Exact Equation: M(x,y) dx + N(x,y) dy = 0 is exact if there exists F(x,y) with:

∂F/∂x = M and ∂F/∂y = N

Exactness Test: The equation is exact if and only if ∂M/∂y = ∂N/∂x (assuming continuous partials).

Solving Procedure

Write the ODE as M dx + N dy = 0.
Check: does ∂M/∂y = ∂N/∂x?
If yes, integrate M with respect to x: F = ∫M dx + g(y).
Differentiate F with respect to y and set equal to N to find g'(y).
Integrate g'(y) to get g(y). The solution is F(x,y) = C.

Example 1: Solve (2xy + 3) dx + (x² + 4y) dy = 0

Step 1: M = 2xy + 3, N = x² + 4y.

Step 2: ∂M/∂y = 2x. ∂N/∂x = 2x. Equal! The equation is exact.

Step 3: F = ∫(2xy + 3) dx = x²y + 3x + g(y).

Step 4: ∂F/∂y = x² + g'(y). Set equal to N: x² + g'(y) = x² + 4y. So g'(y) = 4y.

Step 5: g(y) = 2y².

Solution: x²y + 3x + 2y² = C.

Example 2: Solve (ye^xy + 2x) dx + (xe^xy - 1) dy = 0

M = ye^xy + 2x, N = xe^xy - 1.

∂M/∂y = e^xy + xye^xy. ∂N/∂x = e^xy + xye^xy. Equal -- exact!

F = ∫(ye^xy + 2x) dx = e^xy + x² + g(y).

∂F/∂y = xe^xy + g'(y) = xe^xy - 1. So g'(y) = -1, g(y) = -y.

Solution: e^xy + x² - y = C.

Example 3: Not Exact

Is (y + 1) dx + (2x) dy = 0 exact?

M = y + 1, N = 2x. ∂M/∂y = 1, ∂N/∂x = 2. Not equal -- not exact.

This equation requires an integrating factor or a different method.

Check Your Understanding

1. Test for exactness: (3x² + y) dx + (x + 2y) dy = 0.

∂M/∂y = 1. ∂N/∂x = 1. Exact.

2. Solve the equation from Problem 1.

F = ∫(3x²+y)dx = x³ + xy + g(y). ∂F/∂y = x + g'(y) = x + 2y. So g'(y)=2y, g(y)=y². Solution: x³ + xy + y² = C.

3. Is (2y) dx + (x) dy = 0 exact?

∂M/∂y = 2, ∂N/∂x = 1. Not equal. Not exact.

4. Solve (cos y) dx + (-x sin y + 1) dy = 0.

∂M/∂y = -sin y = ∂N/∂x. Exact. F = ∫cos y dx = x cos y + g(y). ∂F/∂y = -x sin y + g'(y) = -x sin y + 1. g'(y) = 1, g(y) = y. Solution: x cos y + y = C.

5. What does the exactness condition ∂M/∂y = ∂N/∂x mean geometrically?

It means the vector field (M, N) is conservative -- it is the gradient of some potential function F. The solution curves F(x,y) = C are the level curves of F.

Key Takeaways

Write the ODE as M dx + N dy = 0 and check ∂M/∂y = ∂N/∂x.
If exact, find F by integrating M w.r.t. x (or N w.r.t. y), adding an unknown function of the other variable.
Determine the unknown function by comparing with the other partial derivative.
The implicit solution is F(x,y) = C.
If not exact, an integrating factor or different technique is needed.

Next Lesson

Substitution methods for Bernoulli and homogeneous equations.

Lesson 4

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