(sin(y) − Y Sin(x)) Dx + (cos(x) + X Cos(y) − Y) Dy = 0

Jasonbradley

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Jun 16, 2024

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Solving the Differential Equation (sin(y) − y sin(x)) dx + (cos(x) + x cos(y) − y) dy = 0

This differential equation is of the form M(x, y) dx + N(x, y) dy = 0, where:

• M(x, y) = sin(y) − y sin(x)
• N(x, y) = cos(x) + x cos(y) − y

To determine if this equation is exact, we need to check if the following condition holds:

∂M/∂y = ∂N/∂x

Let's calculate the partial derivatives:

• ∂M/∂y = cos(y) - sin(x)
• ∂N/∂x = -sin(x) + cos(y)

Since ∂M/∂y = ∂N/∂x, the given differential equation is exact. This means we can find a potential function Φ(x, y) such that:

• ∂Φ/∂x = M(x, y)
• ∂Φ/∂y = N(x, y)

Let's integrate ∂Φ/∂x = M(x, y) with respect to x:

Φ(x, y) = ∫(sin(y) - y sin(x)) dx = x sin(y) + y cos(x) + h(y)

Here, h(y) is an arbitrary function of y, as the integration was with respect to x.

Now, differentiate Φ(x, y) with respect to y:

∂Φ/∂y = x cos(y) + cos(x) + h'(y)

We know that ∂Φ/∂y = N(x, y). Comparing the two expressions, we get:

x cos(y) + cos(x) + h'(y) = cos(x) + x cos(y) - y

This implies h'(y) = -y. Integrating this with respect to y, we obtain:

h(y) = -y²/2 + C

where C is a constant of integration.

Substituting h(y) back into the expression for Φ(x, y), we get:

Φ(x, y) = x sin(y) + y cos(x) - y²/2 + C

Therefore, the general solution to the given differential equation is:

x sin(y) + y cos(x) - y²/2 = C

where C is an arbitrary constant.