Question

Let 𝑢 = 𝑓(𝑥, 𝑦) = (𝑒^𝑥)𝑠𝑒𝑛(3𝑦). Check if 9((𝜕^2) u / 𝜕(𝑥^2)) +((𝜕^2) 𝑢 / 𝜕(𝑦^2)) = 0

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Frederik

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To check if \(9\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\) , we need to calculate the second partial derivatives of u with respect to x and y, and then verify if the equation holds.
First, let's find the first and second partial derivatives of \(u = f(x, y) = e^x \sin(3y)\) :
1. First partial derivatives:
\(\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(e^x \sin(3y))\)
Using the product rule, we have:
\(\frac{\partial}{\partial x}(e^x \sin(3y)) = e^x \sin(3y) + e^x \frac{\partial}{\partial x}(\sin(3y))\)
\(\frac{\partial u}{\partial x} = e^x \sin(3y) + 0\)
\(\frac{\partial u}{\partial x} = e^x \sin(3y)\)
2. Now, let's find the second partial derivative with respect to x:
\(\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}(e^x \sin(3y))\)
Using the product rule again:
\(\frac{\partial}{\partial x}(e^x \sin(3y)) = e^x \sin(3y) + e^x \frac{\partial}{\partial x}(\sin(3y))\)
The second term, \(\frac{\partial}{\partial x}(\sin(3y))\) , is 0 since the derivative of sin(3y) with respect to x is 0.
So, \(\frac{\partial^2 u}{\partial x^2} = e^x \sin(3y)\)
3. Next, let's find the first partial derivative with respect to y:
\(\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(e^x \sin(3y))\)
Using the chain rule, we have:
\(\frac{\partial}{\partial y}(e^x \sin(3y)) = e^x \cdot 3\cos(3y)\)
\(\frac{\partial u}{\partial y} = 3e^x\cos(3y)\)
4. Finally, let's find the second partial derivative with respect to y:
\(\frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y}(3e^x\cos(3y))\)
Using the chain rule again:
\(\frac{\partial}{\partial y}(3e^x\cos(3y)) = 3e^x\cdot(-3\sin(3y)) = -9e^x\sin(3y)\)
Now, we can plug these results into the equation
\(9\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\) :
\(9(e^x \sin(3y)) + (-9e^x\sin(3y)) = 0\)
\(9e^x \sin(3y) - 9e^x\sin(3y) = 0\)
The equation simplifies to:
\(0 = 0\)
So, the equation \(9\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\) holds, and it is satisfied.

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