(X^2 + y)dx + (x + e^y)dy = 0

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Solve First Order ODE

Solve This First Order Ordinary Differential Equation
(X^2 + y)dx + (x + e^y)DT

To solve the first-order ordinary differential equation (ODE):

(X^2 + y)dx + (x + e^y)dy = 0

We can check if it is exact by verifying if the partial derivatives with respect to x and y are equal. Let's calculate those:

∂/∂y (X^2 + y) = 1
∂/∂x (x + e^y) = 1

Since the partial derivatives are equal, the equation is exact.

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

∂Φ/∂x = X^2 + y
∂Φ/∂y = x + e^y

Integrating the first equation with respect to x:

Φ(x, y) = (1/3)x^3 + xy + f(y)

Here, f(y) is an arbitrary function of y alone. Next, we differentiate Φ(x, y) with respect to y and set it equal to the second equation:

∂Φ/∂y = x + e^y
d/dy [(1/3)x^3 + xy + f(y)] = x + e^y

Differentiating the terms with respect to y:

d/dy (1/3)x^3 + d/dy (xy) + d/dy [f(y)] = x + e^y
0 + x + f'(y) = x + e^y

We can see that f'(y) = e^y. Integrating both sides with respect to y:

∫f'(y) dy = ∫e^y dy
f(y) = e^y + C

Here, C is the constant of integration.

Substituting this value of f(y) into our potential function:

Φ(x, y) = (1/3)x^3 + xy + e^y + C

Now, we have found the potential function Φ(x, y). To find the solution to the exact ODE, we equate Φ(x, y) to a constant, K:

(1/3)x^3 + xy + e^y + C = K

This is the general solution to the given first-order ODE.

Thanks brother 🥰

@chiboyswag

clockwisereport:
@chiboyswag

thank you Sir