Final answer:
The given differential equation is not exact initially. However, when multiplied by the integrating factor \(x\mathrm{e}^{x}\), it becomes exact. One can then solve the modified equation by integrating the resulting M and N with their respective variables.
Step-by-step explanation:
The given differential equation is (x+2)\(\sin y\)+(x\(\cos y\))y'=0. To show whether it is exact, we need to find M and N such that M = (x+2)\(\sin y\) and N = x\cos y\. The exactness condition requires that \(\partial M / \partial y\) is equal to \(\partial N / \partial x\). Differentiating M with respect to y and N with respect to x, we see that they are not equal, implying that the equation is not exact.
When multiplied by the integrating factor \(\mu(x,y)=x\mathrm{e}^{x}\), it becomes exact. Multiplying through, the new M becomes \(x^2\mathrm{e}^{x}(\sin y)\) and the new N becomes \(x^2\mathrm{e}^{x}(\cos y)\). Checking the exactness again, \(\partial (M) / \partial y\) = \(\partial (N) / \partial x\), which means it now satisfies the exactness condition. Then, we can solve the exact differential equation by integrating M with respect to x and N with respect to y to find the solution.