Week 5 5. Partial derivatives ۞ Consider a function of two variables f(x, y). To calculate the partial derivative of f with respect to x, one should differentiate f with respect to x, while treating y as a constant. Notation:

Example 1: Given find fx at (4, –5). Answer: fx = 2x + 3y2 = 2 × 4 + 3 × 52 = 83.

Second-order derivatives: “d squared f d x squared” “d squared f d x d y” The method of calculation:

Example 2: Given find fxx, fyy, fxy, fyx. Solution: Observe that fxy = fyx.

Theorem 1: The Mixed Derivatives Theorem If f(x, y) and its partial derivatives fxx, fyy, fxy and fxy exist and are continuous at a point (x0, y0), then ۞ A linear approximation of a function of a single variable is ۞ A linear approximation of a function of two variables is

۞ An expression of the form is called the differential of the function f(x, y).

6. Exact ODEs Any 1st-order ODE, can be rewritten in the form (1) In some cases, (1) can be further rewritten as (2) where the function f(x, y) is such that

Eq. (2) can be immediately solved in an implicit form: Example 3: The ODE (3) Can be rewritten in the form df = 0, where Hence the solution of (3) is

۞ An ODE of the form (1) is called exact if a function f(x, y) exists such that Q: What happens if we are not given f(x, y)? Can we still test an ODE for exactness?

Theorem 2: If M(x, y) and N(x, y) are smooth (i.e. have continuous derivatives) everywhere in R2, then (1) is exact if and only if Proof consist of two parts: 1. If ODE (1) is exact, then My = Nx. 2. If My = Nx, then ODE (1) exact. We’ll prove only part 1 (part 2 is harder to prove).

If ODE (1) is exact – hence, Then, calculate (4) Due to the Mixed Derivatives Theorem (Theorem 1), fxy = fyx – hence, (4) implies that My = Nx, as required. 

Example 4: (a) isn’t exact because

Example 4 (continued): (b) is exact because

Q: We’ve tested an ODE, and it’s turned out to be exact Q: We’ve tested an ODE, and it’s turned out to be exact. How can we solve it? Example 5: Solve Solution: This ODE has been tested in Example 5b and has turned out to be exact. Hence, there exists f(x, y) such that (5) (6)

Solving (5) (7) where k(y) is a ‘constant’ of integration with respect to x. Substitute (7) into (6): hence, hence,

(8) where the constant of integration was omitted. Substitute (8) into (7): Finally, the solution to the original ODE is Q: What would happen if we kept the constant of integration in (8)?

۞ Consider a non-exact ODE and multiply it by a function I(x, y) ۞ Consider a non-exact ODE and multiply it by a function I(x, y). If after that the ODE becomes exact, I(x, y) is called an integrating factor for this ODE (“an” because can be more than one). Example 6: Show that I = y–2 is an integrating factor for (9) and solve it. Solution: Step 1: Multiply (9) by I, (10) hence,

Testing (10) for exactness: Since (Mnew)y = (Nnew)x , the test’s passed. Step 2: Solving (10): This is NOT the answer The answer is:

Exact ODEs (summary): (11) 1. Eq. (11) is exact if 2. If it indeed is, f(x, y) exists such that (12) 3. Solve (12) for f(x, y) and write down the answer in the form 4. If (11) is not exact...

4. If (11) is not exact (i.e. My ≠ Nx), then it can be made exact by multiplying by a suitable integrating factor I(x, y): Mnew Nnew If I(x, y) is chosen well, then and the ODE has indeed become exact (hence, can be solved through the standard procedure).