Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.

Answer:[tex]2*sin(x)+y*cos(x)-cos(y)=C_1[/tex]Step-by-step explanation:Let:[tex]P(x,y)=2*cos(x)-y*sin(x)[/tex][tex]Q(x,y)=cos(x)+sin(y)[/tex]This is an exact differential equation because:[tex]\frac{\partial P(x,y)}{\partial y} =-sin(x)[/tex][tex]\frac{\partial Q(x,y)}{\partial x}=-sin(x)[/tex]With this in mind let's define f(x,y) such that:[tex]\frac{\partial f(x,y)}{\partial x}=P(x,y)[/tex]and[tex]\frac{\partial f(x,y)}{\partial y}=Q(x,y)[/tex]So, the solution will be given by f(x,y)=C1, C1=arbitrary constantNow, integrate [tex]\frac{\partial f(x,y)}{\partial x}[/tex] with respect to x in order to find f(x,y)[tex]f(x,y)=\int\  2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)[/tex]where g(y) is an arbitrary function of yLet's differentiate f(x,y) with respect to y in order to find g(y):[tex]\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}[/tex]Now, let's replace the previous result into [tex]\frac{\partial f(x,y)}{\partial y}=Q(x,y)[/tex] :[tex]cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)[/tex]Solving for [tex]\frac{dg(y)}{dy}[/tex][tex]\frac{dg(y)}{dy}=sin(y)[/tex]Integrating both sides with respect to y:[tex]g(y)=\int\ sin(y)  \, dy =-cos(y)[/tex]Replacing this result into f(x,y)[tex]f(x,y)=2*sin(x)+y*cos(x)-cos(y)[/tex]Finally the solution is f(x,y)=C1 :[tex]2*sin(x)+y*cos(x)-cos(y)=C_1[/tex]