Final answer:
To show that any separable equation M(x)+N(y)y'=0 is also exact, we need to demonstrate that it satisfies the condition M(x,y)+N(x,y)y'=0. Let's start by assuming that the given equation is separable, which means we can write it as M(x)+N(y)y'=0. We need to show that M(x,y)+N(x,y)y'=0.
Step-by-step explanation:
To show that any separable equation M(x)+N(y)y'=0 is also exact, we need to demonstrate that it satisfies the condition M(x,y)+N(x,y)y'=0. Let's start by assuming that the given equation is separable, which means we can write it as M(x)+N(y)y'=0. We need to show that M(x,y)+N(x,y)y'=0. Since the left-hand sides of the two expressions are both zero, we can set them equal to each other. This gives us M(x)+N(y)y'=M(x,y)+N(x,y)y'. Rearranging the terms, we obtain M(x)-M(x,y)=N(x,y)y'-N(y)y'. Since both sides of the equation are equal to zero, we have M(x)-M(x,y)=0 and N(x,y)y'-N(y)y'=0. This shows that the separable equation is also exact.