### Author Topic: TUT0101  (Read 2256 times)

#### Yiheng Bian

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• Posts: 29
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##### TUT0101
« on: October 05, 2019, 11:39:56 AM »
ANSWER

#### Yiheng Bian

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• Posts: 29
• Karma: 12
##### Re: TUT0101
« Reply #1 on: October 05, 2019, 05:02:20 PM »
$$(x+2)sin(y)+xcos(y)y’=0, \mu(x,y)=xe^x\\$$
$$M=(x+2)cos(y), N=xcos(y)\\$$
$$M_y=(x+2)cos(y), N_x=cos(y)\\$$
$$Since M_y \neq N_x, \text{so the equation is not exact}\\$$
$$\text{Now multiply}\mu=xe^x \text{both sides}\\$$
$$(x+2)xe^xsin(y)+x^2e^xcos(y)y’=0\\$$
$$M’=(x+2)xe^xsin(y), N’=x^2e^xcos(y)\\$$
$$M’_y=cos(y)(x^2e^x+2xe^x), N’_x=cos(y)(2xe^x+x^2e^x)\\$$
$$M’_y=N’_x\\$$
$$Then$$
$$\psi_x=M’=(x+2)xe^xsin(y)=(x^2e^x+2xe^x)sin(y)\\$$
$$\psi=x^2e^xsin(y)+h(y)\\$$
$$\psi=N’=cos(y)x^2e^x+h’(y)=cos(y)x^2e^x\\$$
$$So h’(y)=0\\$$
$$h( y) \text{is constant C}\\$$
$$So \psi=x^2e^xsin(y)+C\\$$
$$So x^2e^xsin(y)=C\\$$

#### Yiheng Bian

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• Posts: 29
• Karma: 12
##### Re: TUT0101
« Reply #2 on: November 18, 2019, 06:16:04 PM »
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