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Quiz-4 / Re: Quiz 4 -- both sections
« on: March 06, 2018, 04:11:59 PM »
Sorry I don't understand why we can assume $\lambda$ to be a positive number ( here $\omega^4$), can someone help?
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Quiz-1 / Re: Q1-T5102-P1
« on: January 25, 2018, 08:48:59 AM »
Since we have $u_{xy}=2u_x$
Let $v = u_x$
Now we have : $v_y = 2v $
which implies $v= \alpha'(x)e^{2y}$
integral with respect to x: $\int v dx = \int \alpha'(x)e^{2y} dx$
$ u = \alpha(x)e^{2y} + \phi(y)$
Check: $u_x = \alpha'(x)e^{2y} + 0$
$u_{xy} = 2\alpha'(x)e^{2y} = 2u_x$
Let $v = u_x$
Now we have : $v_y = 2v $
which implies $v= \alpha'(x)e^{2y}$
integral with respect to x: $\int v dx = \int \alpha'(x)e^{2y} dx$
$ u = \alpha(x)e^{2y} + \phi(y)$
Check: $u_x = \alpha'(x)e^{2y} + 0$
$u_{xy} = 2\alpha'(x)e^{2y} = 2u_x$
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Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 06, 2018, 06:33:04 PM »
Here is my solution to this question:
Given $u_{xx} = 0 \text{ , } u_{yy} = 0$ and $u=u(x,y)$
$u_{xx}=0$ implies $u_x = \phi(y)$
so we have $u(x,y) = \int u_x dx = \phi(y) x + \psi(y) $ (*)
And from $u_{yy}=0$, we have $u_y = \varphi(x)$
With (*) implies $u(x,y)_y = {(\phi(y) x + \psi(y))}_y = \varphi(x)$
Partial derivative on both sides with respect to y we have:
$\phi_{yy}(y)x + \psi_{yy}(y) = 0 $
which implies that:
$\phi_y(y) = c$ and $\psi_y(y) = d$ where both of c and d are constant
Thus:
$\phi(y) = cy + a$ and $\psi(y) = dy + b$ (**) where both of a and b are constant
Combine (*) and (**): we have $u(x,y) = cyx + ax + dy +b$ as our general solution to this pde question
Given $u_{xx} = 0 \text{ , } u_{yy} = 0$ and $u=u(x,y)$
$u_{xx}=0$ implies $u_x = \phi(y)$
so we have $u(x,y) = \int u_x dx = \phi(y) x + \psi(y) $ (*)
And from $u_{yy}=0$, we have $u_y = \varphi(x)$
With (*) implies $u(x,y)_y = {(\phi(y) x + \psi(y))}_y = \varphi(x)$
Partial derivative on both sides with respect to y we have:
$\phi_{yy}(y)x + \psi_{yy}(y) = 0 $
which implies that:
$\phi_y(y) = c$ and $\psi_y(y) = d$ where both of c and d are constant
Thus:
$\phi(y) = cy + a$ and $\psi(y) = dy + b$ (**) where both of a and b are constant
Combine (*) and (**): we have $u(x,y) = cyx + ax + dy +b$ as our general solution to this pde question
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