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### Messages - Ziqian Qiu

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##### Term Test 2 / Re: Problem 4 (noon)
« on: November 19, 2019, 05:30:32 AM »
nvm I just updated the previous post

2
##### Term Test 2 / Re: Problem 4 (noon)
« on: November 19, 2019, 05:24:11 AM »
this is my solution

3
##### Quiz-2 / TUT0202 Quiz2
« on: October 04, 2019, 08:07:06 PM »
want to show $(x+2)sin(y) + x cos (y)y'=0$ is not exact but turns exact after multiplies the given integrating factor $u=xe^x$, then solve it
we let $M = (x+2)sin(y)$ and $N = x cos (y)$
calculate $M_y = (x+2)cos(y)$ and $N_x = cos(y)$
therefore $M_y \neq N_x$ , therefore the equation is not exact.
after multiplies the given integrating factor, the equation becomes $(x+2)xe^xsin(y) + x^2e^x cos (y)y'=0$
this time we let $M = (x+2)xe^xsin(y)$ and  $N = x^2e^x cos (y)$
now we calculate $M_y = (x+2)xe^xcos(y)$ and $N_x = x^2e^xcos(y)$ + $2xe^xcos(y)$
then $M_y = N_x$,  therefore it becomes exact after times the integrating factor.
want to find $\phi (x,y)$ s.t $\phi_x = M$ and $\phi_y = N$
then $\phi = \int x^2e^x cos (y)dy$
the $\phi = x^2e^x sin (y) + h(x)$
then $\phi_x = x^2e^xsin(y)$ + $2xe^xsin(y) + h'(x) = M = (x+2)xe^xsin(y) + 0$
then $h'(x) = 0$
then $h(x) = c$
therefore $\phi(x,y) = x^2e^x sin (y) + h(x) = c$

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