Author Topic: Day section 2.6 #25  (Read 3703 times)

Victor Ivrii

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Day section 2.6 #25
« on: January 21, 2013, 11:05:13 AM »
Please post solution:

Find an integrating factor and solve the given equation
$$
(3x^2y + 2xy + y^3) + (x^2 + y^2)y '= 0.
$$

Changyu Li

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Re: Day section 2.6 #25
« Reply #1 on: January 21, 2013, 02:49:19 PM »
$$
M = 3x^2 y + 2xy + y^3 \\
N = x^2 + y^2 \\
\frac{\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}}{N} \mu = \frac{d\mu}{dt} \\
\frac{3x^2+3y^2}{x^2+y^2} \mu = \frac{d\mu}{dx} \\
$$
new equation is separable
$$
\mu = e^{3x} \\
\frac{\partial \psi}{\partial x} = e^{3x}\left( 3 x^2 y + 2 x y + y^3 \right) \\
\psi = e^{3x} y \left( x^2 + y^2/3 \right) + g(y) \\
\\
\frac{\partial \psi}{\partial y} = e^{3x} \left( x^2 + y^2 \right) \\
\psi = e^{3x} y \left( x^2 + y^2/3 \right) + \tilde{g}(x) \\
\psi = e^{3x} y \left( x^2 + y^2/3 \right) \\
e^{3x} y \left( x^2 + y^2/3 \right) = C
$$
« Last Edit: January 22, 2013, 03:35:49 AM by Victor Ivrii »

Victor Ivrii

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Re: Day section 2.6 #25
« Reply #2 on: January 21, 2013, 03:09:30 PM »
You need to provide more details:
$M=3x^2y+2xy+y^3$, $N=x^2+y^2 \implies R:=\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}=3x^2+3y^2$ and $\frac{R}{N}=3$ is function of $x$ only, then we look for $\mu=\mu(x)$ s.t. $\mu' =3$.

Also after $\frac{\partial \psi}{\partial x}= e^{3x}(3x^2y+2xy+y^3)$ you need to provide transition to $\psi= \ldots$. In fact, much simpler to start from $\frac{\partial \psi}{\partial y}=e^{3x}(x^2+y^2) \implies \psi =e^{3x}(x^2y+ \frac{1}{3}y^3)+h(x)$ and then find $h'(x)=0\implies h(x)=C$.

Changyu Li

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Re: Day section 2.6 #25
« Reply #3 on: January 21, 2013, 11:03:52 PM »
Fixed, I think.