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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

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nvm I just updated the previous post

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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 $\phi(x,y) = x^2e^x sin (y) + h(x) = c$

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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