Answered: "Prove by laplace equation f(x,y)=e−2ycos⁡2x"

Clifland

Answered question

2021-10-12

Prove by laplace equation

f (x, y) = e^{- 2 y} \cos 2 x

Delorenzoz

Skilled2021-10-13Added 91 answers

Step 1
The function given as,

f (x, y) = e^{- 2 y} \cos 2 x

The condition for Laplace equation,

\frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} = 0

.
Take the first partial derivatives of the given function ,

\frac{\partial f}{\partial x} = \frac{\partial (e^{- 2 y} \cos 2 x)}{\partial x}

= - 2 e^{- 2 y} \sin 2 x

\frac{\partial f}{\partial y} = \frac{\partial (e^{- 2 y} \cos 2 x)}{\partial y}

= - 2 e^{- 2 y} \cos 2 x

.
Step 2
The second partial derivatives are,

\frac{\partial^{2} f}{\partial x^{2}} = \frac{\partial}{\partial x} (\frac{\partial f}{\partial x})

= - 4 e^{- 2 y} \cos 2 x

\frac{\partial^{2} f}{\partial y^{2}} = \frac{\partial}{\partial y} (\frac{\partial f}{\partial y})

= 4 e^{- 2 y} \cos 2 x

.
Substitute in the Laplace equation,

\frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} = 0

- 4 e^{- 2 y} \cos 2 x + 4 e^{- 2 y} \cos 2 x = 0

.
Thus, the function is proved by Laplace equation.

Jeffrey Jordon

Expert2022-03-13Added 2605 answers

Answer is given below (on video)

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