Verify that the given differential equation is not exact....

Verify that the given differential equation is not exact. Multiply the given differential equation by the indicated integrating factor $\mu(x, y)$ and verify that the new equation is exact. Solve. $$\begin{array}{l} (-x y \sin x+2 y \cos x) d x+2 x \cos x d y=0 \\ \mu(x, y)=x y \end{array}$$

Verify that the given differential equation is not exact. Multiply the given differential equation by the indicated integrating factor $\mu(x, y)$ and verify that the new equation is exact. Solve.
$$\begin{array}{l}
(-x y \sin x+2 y \cos x) d x+2 x \cos x d y=0 \\
\mu(x, y)=x y
\end{array}$$

Key Concepts

Exact Differential Equations

An exact differential equation is one where the differential expression can be written as the total differential of a function. This means that if an equation is expressed in the form M(x, y) dx + N(x, y) dy = 0, there exists a function F(x, y) such that dF = M dx + N dy. The condition for exactness is that the partial derivative of M with respect to y matches the partial derivative of N with respect to x, ensuring the existence of a potential function F from which the equation can be derived.

Integrating Factors

An integrating factor is a function that, when multiplied by a non-exact differential equation, transforms it into an exact one. Integrating factors depend on one or both of the variables and are chosen such that after multiplication, the new coefficient functions meet the exactness condition. They play a crucial role in solving differential equations that are not initially exact.

Condition for Exactness

The condition for a differential equation to be exact is given by the equality of the mixed partial derivatives: ?M/?y = ?N/?x, where M and N are the coefficients of dx and dy respectively. This condition guarantees that there exists a function F(x, y) with continuous second partial derivatives, confirming that F_x equals M and F_y equals N.

Solving Exact Differential Equations

Once a differential equation is verified to be exact, or made exact by an integrating factor, the next step is to find the potential function F(x, y). This is done by integrating one of the functions (M or N) with respect to its variable while treating the other variable as a constant and then determining the arbitrary function of the other variable. The final solution is given by setting F(x, y) equal to a constant, which represents the implicit solution of the differential equation.

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