Sometimes a nonexact differential equation M(x,y)dx + N(x,y)dy = 0 can be made exact by multiplying the equation by another function, say Î¼(x,y). In other words, Î¼Mdx + Î¼Ndy = 0 becomes exact. We have a such a case as I will show shortly after through examples. In this case, we call the function Î¼(x,y) an integrating factor of the given nonexact first order differential equations say, Mdx + Ndy = 0. Now think about the following example. I'm claiming that the differential equation (6) where the M is equal to negative xy sin x + 2y cos x and N is equal to 2x cos x. Take x a partial of the y partial derivative of M, and x partial derivative of N, and see whether they are equal or not. You can easily see that they do not equal. That means the given differential equation (6), this equation, this is not exact. But my claim, multiply this equation by Î¼(x,y) which is equal to xy. Then it becomes exact. In other words, the Î¼(x,y) is equal to x times y is an integrating factor of this differential equation (6). So check it, and then solve the equation. That's the problem. As I said, it's very easy to see that the equation (6)'s surface not exact. Please confirm it. Now multiply the given differential equation (6) by xy. Then it gives you -xÂ²yÂ² sin x + 2xyÂ² cos x dx + 2xÂ² y cos x dy = 0. Now we have a new M down here, and a new N down here. Compute to the dM/dy. Compute the dM/dy then from this first part by the product flow, you have dM/dy, -2xÂ²y sin x + 4xy cos x. On the other hand, the dN/dx is equal to, from this one, by the product flow, you have 4xy cos x, then 2xÂ²y times derivative of cos x, that is equal to negative sin of x. In total you get, -2xÂ²y times the sin x. dM/dy is the same as dN/dx. What does that mean? This new differential equation which we obtain by multiplying the equation (6) by the integrating factor xy, this is really exact. In other words, Î¼(x,y) is equal to x times y, this is really an integrating factor of the equation (6). Now, we are ready to solve this exact differential equation. There must be a function F, whose y partial derivative is equal to N of xy, which is 2xÂ²y and cos x. Then, what is the F? Through the integration, F is equal to integral of 2xÂ²y cos xdy plus integrating constant, which is an arbitrary differential function of x that I denoted by g(x). Now the integration will be 2xÂ² times the cos x is a constant respect to the dy integration. You simply have xÂ² and yÂ² cos x. That's the antiderivative of this. Our candidate for F will be xÂ² and yÂ² cos x + g(x). This candidate must satisfied another conditions say, dF/dx = M. First, from this the expression here, what is dF/dx? From this part of product flow, it's first 2xyÂ² cos x then plus xÂ² times yÂ² times derivative of cos x is a negative sign of x. So here you get the negative xÂ²yÂ² sin x. Finally, derivative of g will be g' of x. That must be equal to M. M is of this one. Compare these two. First, negative xÂ²yÂ² sin x, it's the same as here. They canceled out and 2xyÂ² cos x, they are common in both sides, so they canceled out again. We get g'(x) is identically zero. That means we can take g(x) to be zero. Finally, we get xÂ²yÂ² cos x and plus zero, that is equal to arbitrary constant c down there. This is a general solution of the given nonexact differential equation (6). Remember that we find the general solution, by first checking out that the function x times y is an integrating factor of the differential equation (6).