### Author Topic: Q2-T5101  (Read 2398 times)

#### Victor Ivrii ##### 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

• Newbie
• • Posts: 1
• Karma: 0 ##### 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

• Full Member
•   • Posts: 27
• Karma: 29 ##### 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 »

#### Victor Ivrii ##### Re: Q2-T5101
« Reply #3 on: February 02, 2018, 06:24:31 PM »
ruantian
Please select a proper screen name. Second, your solution is correct but the answer is not due to the lame method to write. Do not use html tags for math, they are inadequate. Use MathJax

Junya
Do not post solutions after correct one was posted. It would be much better if you wrote something like:  "ruantian solved problem correctly, but presented an answer in the confusing way, correct answer is ... "