### Author Topic: Q2-T5101  (Read 2413 times)

#### Victor Ivrii

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##### Q2-T5101
« on: February 02, 2018, 02:15:17 PM »
Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$x^2y^3 + x(1 + y^2)y' = 0,\qquad \mu(x, y) = 1/xy^3.$$

#### Tianjing Ruan

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##### Re: Q2-T5101
« Reply #1 on: February 02, 2018, 02:59:04 PM »
x2y3/xy3 + x(1 + y2)/xy3y' = 0

x + (y−3 + y−1)y' = 0

∫M dx = 1/2x2 + g(y)

g(y) = ∫y−3 + 1/ydy = -1/2y−2 + ln(y)

The solution is:
1/2x2 − 1/2y2 + ln(y) = C

#### Junya Zhang

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##### Re: Q2-T5101
« Reply #2 on: February 02, 2018, 05:56:29 PM »
First, let's show the given DE $x^{2}y^{3} + x(1+y^{2})y' = 0$ is not exact.
Define $M(x,y)=x^{2}y^{3}$, $N(x,y)=x(1+y^{2})$
$$M_y = \frac{\partial}{\partial y}[x^{2}y^{3}] = 3x^{2}y^{2}$$ $$N_x = \frac{\partial}{\partial x}[x(1+y^{2})] = 1+y^{2}$$
Since $3x^{2}y^{2} ≠ 1+y^{2}$, this implies the given DE is not exact.

Now, let's show that the given DE multiplied by the integrating factor $\mu(x,y) = \frac{1}{xy^{3}}$ is exact.
That is to show $$\frac{1}{xy^{3}}x^{2}y^{3} + \frac{1}{xy^{3}}x(1+y^{2})y' = x + (y^{-3}+y^{-1})y'= 0$$ is exact.

Define $M'(x,y) = x$, $N'(x,y) = y^{-3}+y^{-1}$
Since
$$M'_y = \frac{\partial}{\partial y}(x) = 0$$ $$N'_x = \frac{\partial}{\partial x}[y^{-3}+y^{-1}] = 0$$
By theorem in the book, we can conclude that $x + (y^{-3}+y^{-1})y'= 0$ is exact.

Thus, we know there exists a function $\phi(x,y)=C$ which satisfies the given DE.
Also,
$$\frac{\partial \phi}{\partial x} = x$$ $$\frac{\partial\phi}{\partial y} = y^{-3}+y^{-1}$$
Integrate $\frac{\partial \phi}{\partial x} = x$ with respect to $x$ we have
$$\phi(x,y) = \frac{1}{2}x^{2} + g(y)$$
Take derivative on both sides with respect to $y$ we get
$$\frac{\partial\phi}{\partial y} = g'(y)$$
Since we know that $\frac{\partial\phi}{\partial y} = y^{-3}+y^{-1}$
Then $g'(y) = y^{-3}+y^{-1}$
Integrate with respect to $y$ we have
$g(y) = -\frac{1}{2}y^{-2} + ln|y| + C$
Altogether, we have $\phi(x,y) = \frac{1}{2}x^{2} -\frac{1}{2}y^{-2} + ln|y| = C$, which means
$$\frac{1}{2}x^{2} -\frac{1}{2}y^{-2} + ln|y| = C$$
is the general solution to the given DE.

Besides, notice that the constant function $y(x)=0$ $\forall x$ is also a solution to the given DE.
« Last Edit: February 02, 2018, 06:09:58 PM by Junya Zhang »