Q. 59 - Calculus

Question

For each pair of functions in Exercises 59–62, use Theorem 12.24 to show that there is a function of two variables, $$F(x, y)$$, such that $$\frac{\partial F}{\partial x} =g(x,y)$$ and $$\frac{\partial F}{\partial y} =h(x,y)$$ . Then find F.

$$g(x,y)=e^{x}cosy, h(x,y)=-e^{x}siny+2y$$

Step-by-Step Solution

Answer

The required function is$$F(x,y)=e^{x}cosy+y^{2}+C$$ 

Step 1. Given Information

$$\frac{\partial F}{\partial x} =g(x,y)$$ and $$\frac{\partial F}{\partial y}=h(x,y)$$ 

$$g(x,y)=e^{x}cosy, h(x,y)=-e^{x}siny+2y$$

Step 2. Explanation

Here, we get $$\frac{\partial F}{\partial x}=e^{x}cosy$$ and $$\frac{\partial F}{\partial y}= -e^{x}siny+2y$$

Integrating and evaluating in both the cases, we get

$$F(x,y)=\int e^{x}cosydx$$

$$\implies F(x,y)=e^{x}cosy+P(x)$$   ----(1)

And, $$F(x,y)=\int (e^{x}siny+2y)dy$$

$$\implies F(x,y)=e^{x}siny+y^{2}+C$$   ---(2)

From (1) and (2), we get

$$P(x)=y^{2}$$

Therefore, $$F(x,y)=e^{x}cosy+y^{2}+C$$