91Ó°ÊÓ

A partial derivative is a derivative where we hold one variable constant and differentiate with respect to another.

In the context of the exact differential, we have two functions, M(x,y) and N(x,y), where:

• M(x,y) = \(\frac{x}{x^2 - y^2}\)
• N(x,y) = -\(\frac{y}{x^2 - y^2}\)

To verify if our given differential \(dz = \frac{x dx}{x^2 - y^2} - \frac{y dy}{x^2 - y^2}\) is exact, we check the condition:

\(\frac{\text{d} M}{\text{d} y} = \frac{\text{d} N}{\text{d} x}\). Let's find these partial derivatives:

1. Compute \(\frac{\text{d} M}{\text{d} y}\):

\(\frac{\text{d}}{\text{d} y} \bigg(\frac{x}{x^2 - y^2}\bigg) = \frac{2xy}{(x^2 - y^2)^2}\).

2. Compute \(\frac{\text{d} N}{\text{d} x}\):

\(\frac{\text{d}}{\text{d} x} \bigg( -\frac{y}{x^2 - y^2}\bigg) = \frac{2xy}{(x^2 - y^2)^2}\).

Since these partial derivatives are equal, we confirm the differential equation is exact.

Integration techniques are crucial for solving differential equations. Here, we need to integrate \(\frac{x}{x^2 - y^2}\) with respect to x.

To do this, we use substitution:

• Let \(u = x^2 - y^2\)
• Then, \(\text{d} u = 2x \text{d} x\)

Now, replace \(\text{d} x\) with \(\frac{\text{d} u}{2x}\):

\[ \text{d} z = \frac{x \text{d} u}{2x \times u} = \frac{\text{d} u}{2u} \] Now, integrate:

\[ z = \frac{1}{2} \text{ln}|u| + C(x, y) \] Substitute \(u\) back:

\( z = \frac{1}{2} \text{ln}|x^2 - y^2| + C(y) \).

We are integrating with respect to x, so \(C(y)\) is a function of y only.

Differential equations involve functions and their derivatives. Our specific problem deals with exact differential equations.

An exact differential equation has the form: \( M(x, y) \text{d} x + N(x, y) \text{d} y = 0 \), where \( \frac{\text{d} M}{\text{d} y} = \frac{\text{d} N}{\text{d} x} \).

Here's how we solve it:

• Identify M(x,y) and N(x,y).
• Verify if \( \frac{\text{d} M}{\text{d} y} = \frac{\text{d} N}{\text{d} x} \).
• Integrate M(x,y) with respect to x, and find the constant function C(y).
• Differentiate the resulting function with respect to y, set it equal to N(x,y), and solve for C(y).
• The final solution combines these parts.

For our example:

• M(x, y) = \(\frac{x}{x^2 - y^2}\)
• N(x, y) = -\(\frac{y}{x^2 - y^2}\)

We showed this is exact, integrated M with respect to x, and found the function \(z = \frac{1}{2} \text{ln}|x^2 - y^2| + C(y)\), ultimately confirming \(C(y) = C\), a constant. Finally, we computed the difference between the values of z at two points.