Question

Solve the DE by finding IF
{(x^3)(y^2)-y}dx+{(x^2)(y^4)-x}dy=0

Answer 1

@JamesJ @satellite73 @amistre64 try it plz

Answer 2

you had a question like this yesterday no?

Answer 3

yup but still im having difficulty figuring it out :|

Answer 4

post a link to that other question will you.

Answer 5

http://openstudy.com/users/wasiqss#/updates/4f5c96fce4b0602be4384f7e

Answer 6

Where are you stuck?

Answer 7

w8 imma tell u in min.

Answer 8

james stuck in integration of\[\int\limits \left\{ (1)/\left[ (x^4)y-(x^2)(y^4) \right] \right\}\]

Answer 9

i can suppose x^2 to b X

Answer 10

that is not an integral required in the solution. Where is that in the solution worked out on your other thread?

Answer 11

im doing this one and after applying rule 3, i have got this one , rest portion i have integrated and solved

Answer 12

is this an exact diffy q by chance?

Answer 13

yup amistre it is

Answer 14

whew!! is was hoping it would be simpler than it looked lol

Answer 15

\[(x^3y^2-y)\ dx+(x^2y^4-x)\ dy=0\]
integrate one side and derive it down to the other; then compare notes

Answer 16

and the other integral that giving me headache is \[\int\limits \left\{ 1/((x^3)(y^2)-x(y^5)) \right\}\]

Answer 17

amistre we have to do only by finding integrating factor of it :/

Answer 18

\[\int (x^3y^2-y)\ dx\ = \frac{1}{4}x^4y^2-yx+g(y)\]

\[Dy[\frac{1}{4}x^4y^2-yx+g(y)]=\frac{1}{2}x^4y-x+g'(y)\]

Answer 19

integrating factors eh ....

Answer 20

wasiqss, read your other thread, you were giving the integrating factor, right there!!

Answer 21

james i cant get u

Answer 22

id read the other thread them cause I got no clue how to do this with integrating factors :)

Answer 23

amistre so not right :(

Answer 24

http://openstudy.com/users/wasiqss#/updates/4f5c96fce4b0602be4384f7e

Answer 25

just playing around with it ...
\[\frac{1}{2}x^4y-x+g'(y)=x^2y^4-x\]
\[g'(y)=x^2y^4-x-\frac{1}{2}x^4y+x\]
\[\int g'(y)dy=\int x^2y^4-\frac{1}{2}x^4y\ dy \]
\[g(y) =\frac{1}{5}x^2y^5-\frac{1}{4}x^4y^2+C \]
im sure i did something wrong in there :/

Answer 26

http://www.wolframalpha.com/input/?i=%28x%5E3y%5E2-y%29dx%2B%28x%5E2y%5E4-x%29dy%3D0

oww my eyes!!

Answer 27

amistre u need some brain storming as to how to solve these de's :P

Answer 28

i do :) i do :)

Answer 29

@TuringTest u gonna do?

Answer 30

I've never solved a exact DE problem like this the way ash did on your other post
If I were to try this I would just be trying to imitate his process, which you should be equally capable of
ask me a specific question about some part of the problem and perhaps I could help you

Answer 31

ok help me in the integral

Answer 32

which one?

Answer 33

\[\int\limits \left\{ 1/((x^3)(y^2)-x(y^5)) \right\}\]this one?

Answer 34

yup

Answer 35

\[\int{dx\over x^3y^2-xy^5}\]with respect to x, right?

Answer 36

hello?

Answer 37

@wasiqss

Answer 38

yup

Answer 39

ok, factor out the bottom as much as possible, treating y as constant

Answer 40

w8 w8

Answer 41

ok now what

Answer 42

1/(xy^2)(x^2-y^3)

Answer 43

partial fractions

Answer 44

cause of the y's i get confused in partial fraction , can u help me a bit in that

Answer 45

you can pull 1/y^2 out of the integral first too

Answer 46

so what remains is
1/(x)(x^2-y^3)

Answer 47

yes, and you can do either
A/x+(Bx+C)/(x^2-y^2)
or I suppose use difference of squares and make it
A/x+B/(x+y^(3/2))+C/(x-y^(3/2))

Answer 48

and the whole expression is equal to 1/y rite?

Answer 49

no, once you multiple through by the LCM of the fractions, what's left over will be equal to 1

Answer 50

so this expression A/x+(Bx+C)/(x^2-y^2) is equal to 1?

Answer 51

and 1/y is constant that we will multiply after integration ?

Answer 52

yes, we have to put the 1/y back in the end
we are lust looking at the PF right now\[{A\over x}+{B\over x+y^{3/2}}+{C\over x-y^{3/2}}\]multiply through by \(x(x^2-y^3)\) to get rid of the fractions and what you get will be equal to 1

Answer 53

\[A(x^2-y^3)+Bx(x-y^{3/2})+Cx(x+y^{3/2})=1\]

Answer 54

I do not feel like solving for A, B, and C, so that is going to be your job ;)

Answer 55

ok I'll give you A

Answer 56

let x=0

Answer 57

thanks mate , know can u help with last integral

Answer 58

i did this one :)

Answer 59

cool
I have to go brother, but I'll be seeing you around I'm sure!

Answer 60

cant u help in another one

Answer 61

Sorry, I have class to take and give today, so time for me to prepare
next time friend

Answer 62

okay love u bro :)

Answer 63

lol, 'love' is a strong word, but the platonic feeling is mutual XD

Answer 64

hehehe dont think too much about it cause i could not control my emotions because of u helping me :)

Answer 65

ash do the way u did other day

Answer 66

ash taking long time LOL

Answer 67

Wasiqss do you want me to solve the complete or just the integration?

Answer 68

solving complete would be world of help to me!

Answer 69

u doing?

Answer 70

Wait for some time. I'll have dinner and then I'd solve it here:)

Answer 71

ok thanks but remember to solve :)

Answer 72

yeah

Answer 73

thanks :)

Answer 74

I'll just solve the integral
\[\int \frac{dx}{x^3y^2-xy^5} dx\]
Let's multiply numerator and denominator by \(\large{x^{-3}}\)
we'll get
\[\int \frac{x^{-3}}{y^2-x^{-2}y^5}dx\]
Let
\[y^2-x^{-2}y^5=t\]
so
\[+2x^{-3}y^5dx=dt\]
or
\[x^{-3}dx=\frac{dt}{2y^5}dt\]
we'll get
\[\int \frac{\frac{dt}{2y^5}}{t}dt\]
or
\[\int \frac{dt}{{2y^5}t}dt\]
We get
\[\frac{\ln t}{2y^5}\]
substitute value of t
y^2-x^{-2}y^5
Finally we get
\[\frac{\ln ({y^2-x^{-2}y^5)}}{2y^5}\]

Answer 75

last integral man!!
(1)/((x^4)y-(x^2)(y^4)

Answer 76

We have
\[\int \frac{1}{x^4 y-x^2 y^4}dx\]
Now we can write this as
\[\int \frac{1}{x^2y(x^2 - y^3)}dx\]
Let \[x=y^{3/2} sec \theta \]
\[dx=y^{3/2} \sec \theta \tan \theta d\theta\]
We get
\[\int \frac{y^{3/2} \sec \theta \tan \theta}{y^3 sec^2 \theta \times y(y^3 \sec^2 \theta - y^3)}d \theta\]
W\frac{ e get
\[y^{-11/2} \int \frac{\sec \theta \tan \theta} {\sec ^2 \theta \times \tan^2 \theta} d\theta \]
we get
\[y^{-11/2} \int \frac{cos^2 \theta}{\sin \theta} d\theta\]
We get
\[y^{-11/2} (\int (cosec \theta- \sin \theta )d \theta\]
I think you can do now:)

Answer 77

ash any alternative way , i mean by completing squate of PF?

Answer 78

We have
\[\frac{1}{y} \int \frac{1}{x^2(x^2-y^3)}dx\]
Now rewriting this as
\[\frac{1}{y^4} \int \frac{y^3}{x^2(x^2-y^3)}dx\]
\[\frac{1}{y^4} \int \frac{x^2-(x^2-y^3)}{x^2(x^2-y^3)}dx\]
We get now
\[\frac{1}{y^4} \int (\frac{1}{x^2-y^3}-\frac{1}{ x^2} )dx\]
Now we get
\[\frac{1}{y^4} \int (\frac{1}{x^2-(y^{3/2})^2}-\frac{1}{ x^2} )dx\]
Can you solve it now @wasiqss?

Answer 79

man how u became so good at integration?

Answer 80

@wasiqss that last integral is almost identical to the one I solved with you above
(though I like ash's way better)