Solving the Exact Differential Equation (x-2e^y , dy y , x , sin x , dx 0)

An exact differential equation is a type of first-order differential equation that possesses certain properties that make it easier to solve. Let's solve the equation (x - 2e^y , dy y , x , sin x , dx 0), step by step, using the method of exact differentials.

Step 1: Confirm the Equation is Exact

The given equation is (x - 2e^y , dy y , x , sin x , dx 0). For this to be an exact differential equation, we can write:

[M(x, y) , dx N(x, y) , dy 0]

Here, we identify:

[M(x, y) y , x , sin x quad text{and} quad N(x, y) x - 2e^y]

To confirm exactness, we need to check if (frac{partial M}{partial y} frac{partial N}{partial x}).

[ frac{partial M}{partial y} x sin x quad text{and} quad frac{partial N}{partial x} 1 ]

Since (frac{partial M}{partial y} 1 frac{partial N}{partial x}), the equation is exact.

Step 2: Find the Potential Function (F(x, y))

For an exact differential equation, there exists a function (F(x, y)) such that:

[frac{partial F}{partial x} M(x, y) quad text{and} quad frac{partial F}{partial y} N(x, y)]

We can integrate (M(x, y)) with respect to (x), treating (y) as a constant:

[ F(x, y) int (y , x , sin x) , dx xy , sin x - x , cos x g(y) ]

Next, we integrate (N(x, y)) with respect to (y), treating (x) as a constant:

[ F(x, y) int (x - 2e^y) , dy xy - 2e^y h(x) ]

By comparing the two expressions for (F(x, y)), we can see that:

[g(y) -2e^y quad text{and} quad h(x) -x , cos x]

Therefore, the potential function (F(x, y)) is:

[ F(x, y) xy , sin x - x , cos x - 2e^y]

Step 3: Write the General Solution

The general solution to the exact differential equation is given by:

[ F(x, y) C ]

Therefore, the solution to the given differential equation is:

[xy , sin x - x , cos x - 2e^y C]

Step 4: Verify the Solution

To confirm our solution, we can take the partial derivatives and verify that they satisfy the original differential equation:

[ frac{partial F}{partial x} y , sin x x , cos x - cos x sin x , x y , x , sin x ]

[ frac{partial F}{partial y} x - 2e^y ]

The partial derivatives are correct, and the solution is verified.

Conclusion

The solution to the exact differential equation (x - 2e^y , dy y , x , sin x , dx 0) is:

[ xy , sin x - x , cos x - 2e^y C]

This approach is based on the method of integrating factors and integrating the partial derivatives to find the potential function.