### Author Topic: Re: Q1, P2--Day section  (Read 2783 times)

#### Qian Li

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##### Re: Q1, P2--Day section
« on: October 05, 2013, 03:40:39 PM »
3 from section 2.6, i.e.
$$(3x^2-2xy+2)+(6y^2-x^2+3)y'=0$$
We already know it is exact, partial derivative of $(3x^2-2xy+2)$ with respect to $y$ is $-2x$, while partial derivative of
$(6y^2-x^2+3)$ with respect to $x$ is also $-2x$.

So just assume the original function is  $F(x,y)$. $P(x,y)=3x^2-2xy+2$ is obtained from $F(x,y)$ by taking patial derivative with respect to $x$, and $Q(x,y)=(6y^2-x^2+3)$ with respect to $y$.

Now integrate $P(x,y)$ with respect to $x$, and we can get $F(x,y)=x^3-x^2y+2x+h(y)$. Take partial derivative with respect to $y$, and we get $Q(x,y)= -x^2+h'(y)$. ERROR!!!

If compare this $Q(x,y)$ with the $6y^2-x^2+3$,  we get $h'(y)=-6y^2+3$, which means $h(y)=2y^3+3y+C$, where $C$ is any arbitrary constant.

In conclusion, $F(x,y)=x^3-x^2y+2x+2y^3+3y+C$, or we can write it as $x^3-x^2y+2x+2y^3+3y=C$.

Do not hijack topics--this was about Q1, P1--so I split it
« Last Edit: October 05, 2013, 08:07:30 PM by Qian Li »