Question

(4xy^2+6y)dx+(5yx^2+8x)dy=0
solve by using integrating factor

Answer 1

sam got clue?

Answer 2

(4xy^2+6y)+(5yx^2+8x)dy/dx=0

Answer 3

sam whats in that?

Answer 4

\[y(4xy+6) dx+x(5xy+8)dy=0\]
M=y(4xy+6)
N=x(5xy+8)
Integrating factor=\(\frac{1}{Mx-Ny}\)
Mx=xy(4xy+6)
Ny=xy(5xy+8)
Integrating factor=\[\frac{1}{xy(4xy+6-(5xy+8))}\]
IF=\[\frac{1}{-2xy-x^2y^2}\]
Multiply this by our DE ,
\[\frac{y(4xy+6)}{2xy+x^2y^2} dx+\frac{x(5xy+8)}{2xy+x^2y^2}dy=0\]
First term is M and second is N
solution
\[\int M (y\ treated\ as\ constant)+\int( terms\ of\ N\ not\ containing\ x)dy=C\]
,

Answer 5

hold on till i integrate

Answer 6

ash having hard time integrating it :/

Answer 7

Yeah
After multiplying the integrating factor we get
\[\frac{(4xy+6)y}{xy(xy+2)} dx+\frac{(5xy+8)x}{xy(xy+2)} dy=0\]
We'll get now
\[(\frac{4y}{xy+2}+\frac{6}{x^2y+2x})dx+(\frac{5x}{2+xy}+\frac{8}{y(xy+2)})dy\]
The term with dx we have to treat y as constant and integrate it
In the second term we have to integrate only the terms without x, it may seem taht there is no term without x but we can get a term without x
\[(\frac{4y}{xy+2}+\frac{6/y}{x^2+2x/y})dx+(\frac{5x}{2+xy}+\frac{4(2+xy-xy)}{y(xy+2)})dy\]
\[(\frac{4y}{xy+2}+\frac{6/y}{x^2+2x/y+1/y^2-1/y^2})dx+(\frac{5x}{2+xy}+\frac{8}{y}-\frac{4xy}{y(xy+2)})dy\]
Now Let's integrate
\[4y \ln(xy+2) \times 1/y+ \int \frac{6/y}{(x+1/y)^2-(1/y)^2}dx+ \int \frac{8}{y} dy=c\]
Wasiqq do the integration of middle term, I don't remember its formula and your DE is solve

Answer 8

Can you do it now? Did you understand?

Answer 9

im tryin to understand it

Answer 10

hey jus temme how to integrate 1/(yx^2+2x)

Answer 11

lalaly i want an easy way out to do it

Answer 12

ok i need 10 min...

Answer 13

ok time alloted :P

Answer 14

its 20 min now :P

Answer 15

@lalaly

Answer 16

What is written above is right. There isn't an "easy" way. This is the way.

Answer 17

james im having difficulty integrating it :(

Answer 18

@JamesJ

Answer 19

In that penultimate expression
\[ \int \frac{6/y}{(x + 1/y)^2 - (1/y)^2} dx \]
treat all of the y and 1/y as constants.
This will appear less intimidating if you write a = 1/y
and change variables to x' = x + 1/y = x + a.

Now as dx' = dx, we have
\[ \int \frac{6a}{x'^2 - a^2} dx' \]

Answer 20

@JamesJ final thing how to integrate 1/(yx^2+2x)

Answer 21

withe respect to what? y or x?

Answer 22

with respect to x

Answer 23

taking Y constant

Answer 24

@JamesJ

Answer 25

then complete the square.
yx^2 + 2x = y ( x^2 + (2/y)x )
= y [ ( x^2 + (2/y)x + (1/y)^2 ) - (1/y)^2 ]
= y [ ( x + (1/y) )^2 - (1/y)^2 ]

Answer 26

james look at my other ques