Question

# Let π’ = π$π₯, π¦$ = $π^π₯$π ππ$3π¦$. Check if 9$(π^2$ u / π$π₯^2$) +$(π^2$ π’ / π$π¦^2$) = 0

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## Answer to a math question Let π’ = π$π₯, π¦$ = $π^π₯$π ππ$3π¦$. Check if 9$(π^2$ u / π$π₯^2$) +$(π^2$ π’ / π$π¦^2$) = 0

Frederik
4.6
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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