Show that the following ODE is exact and solve it : (y-x3)dx+(x+y3)dy=0

To determine whether the given ordinary differential equation (ODE) is exact and to solve it, we start with the equation itself: (y - x³)dx + (x + y³)dy = 0. An ODE of the form M(x, y)dx + N(x, y)dy = 0 is exact if the partial derivative of M with respect to y is equal to the partial derivative of N with respect to x. Let's identify M and N first:

Identifying M and N

From the equation, we can define:

• M(x, y) = y - x³
• N(x, y) = x + y³

Checking for Exactness

Next, we need to compute the partial derivatives:

• ∂M/∂y = ∂(y - x³)/∂y = 1
• ∂N/∂x = ∂(x + y³)/∂x = 1

Since ∂M/∂y = ∂N/∂x, we conclude that the ODE is indeed exact.

Finding the Solution

To solve the exact ODE, we need to find a function Ψ(x, y) such that:

• ∂Ψ/∂x = M(x, y)
• ∂Ψ/∂y = N(x, y)

Integrating M with respect to x

We start by integrating M with respect to x:

Ψ(x, y) = ∫(y - x³)dx = yx - (x⁴/4) + h(y),

where h(y) is an arbitrary function of y that arises because the integration is with respect to x.

Finding h(y)

Next, we differentiate Ψ with respect to y:

∂Ψ/∂y = x + h'(y).

Setting this equal to N, we have:

x + h'(y) = x + y³.

From this, we can deduce that h'(y) = y³, which we can integrate to find h(y):

h(y) = (y⁴/4) + C,

where C is a constant of integration.

Combining Results

Now we can substitute h(y) back into Ψ:

Ψ(x, y) = yx - (x⁴/4) + (y⁴/4) + C.

Setting the Level Curve

To find the solution to the ODE, we set Ψ(x, y) equal to a constant:

yx - (x⁴/4) + (y⁴/4) = C.

Final Form of the Solution

This equation represents the implicit solution to the original ODE. It describes a family of curves in the xy-plane, each corresponding to a different value of the constant C. Thus, the solution to the given exact ODE is:

yx - (x⁴/4) + (y⁴/4) = C.

In summary, we verified that the ODE is exact, found the potential function Ψ, and expressed the solution in an implicit form. This method is a powerful tool for solving exact differential equations, and understanding it can greatly enhance your problem-solving skills in differential equations.