-%E2%88%92-y-sin(x))-dx-%2B-(cos(x)-%2B-x-cos(y)-%E2%88%92-y)-dy-%3D-0.jpeg "(sin(y) − Y Sin(x)) Dx + (cos(x) + X Cos(y) − Y) Dy = 0")
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.