Question

Solve the Initial Value Problem

Answer 1

\[x^{2}\frac{ dy }{ dx}=\frac{ 4x^{2}-x-2 }{ (x+1)(y+1) } ,y(1)=1\]

Answer 2

please write step by step, im sorry

Answer 3

\[(y+1) dy =\frac{4x^2-x-2}{x^2(x+1)}dx \]
Ok?

Answer 4

i get that

Answer 5

integral both sides. That's it

Answer 6

okay please bear with me, as it can get crazy

Answer 7

im working on it

Answer 8

I guess so, but that the steps
after getting the answer, replace initial value to get value of C, then plug back

Answer 9

okay give me 10000 seconds please

Answer 10

\[\int\limits_{}^{}\frac{ 4x ^{2}-x-2 }{ x ^{2}(x+1) }dx\]

Answer 11

it has already gotten crazy

Answer 12

use partial fraction, friend

Answer 13

could you do a quick refresher, my apologies as it has been last last year doing this

Answer 14

\[\frac{4x^2-x-2}{x^2(x+1)}=\frac{A}{x+1}+\frac{B}{x}+\frac{C}{x^2}\]
solve for A, B, C you have A = 3, B =1, C =-2
so, the integral becomes
\[\int( \frac{3}{x+1}+\frac{1}{x}-\frac{2}{x^2})dx\]
then take integral each term to get the answer

Answer 15

light BULB! thank you!

Answer 16

ok

Answer 17

im sorry for making you work that out

Answer 18

np.

Answer 19

nevermind!

Answer 20

how do you integrate the y part?

Answer 21

do as usual, = \(\dfrac{y^2}{2}+y\)

Answer 22

i got 3ln(x+1)+lnx+2/x for the other side thank you

Answer 23

+ C

Answer 24

so i got
y^2x/2+y=3ln(x+1)+lnx+2/x+c

Answer 25

yes, I think so

Answer 26

no, y^2 /2, not y^2x/2

Answer 27

you're right my bad

Answer 28

and then solve for y?

Answer 29

solve for C

Answer 30

replace initial value y(1) =1 to solve for c

Answer 31

so plug 1 in for y? im sorry

Answer 32

first off, you have to solve for y . I am sorry, it's not easy but we have an outlet there
y^2 + 2y - 2( the whole thing from the left hand side) =0 . This is a quadratic equation, you solve for y,
so, y = ....( 2 values) then, plug x =1 into this y and the whole thing = 1 because y(1) =1 to solve for c

Answer 33

because from the differential equation above, your answer is y^2/2 +y , not just y, So, you have to solve for y , and then calculate the initial value base on the latest y

Answer 34

gotta go. be back later.

Answer 35

thanks

Answer 36

hey, you see, wolfram solves this problem exactly our way, hehehe
http://www.wolframalpha.com/input/?i=x^2%28dy%2Fdx%29%3D%284x^2-x-2%29%2F%28%28x%2B1%29%28y%2B1%29%29

Answer 37

how does wolfram jump from x^2(dy/dx)=(4x^2-x-2)/((x+1)(y+1) to the long equation with y(x)?

Answer 38

don't you scrool up? it solves the problem at the beginning of the problem, drag the bar up to see the very first step

Answer 39

i don't have an account for that

Answer 40

i don't see it

Answer 41

can you read it?

Answer 42

yes i can

Answer 43

but how do they get from 1/2 y^2+y=3ln(x+1)+lnx+2/x+c to the longer equation with y(x)

Answer 44

hey, don't be panicked.
I told you , it's a quadratic , see -1 at the end of each y
the equation is \[y^2 +2y -(6ln(x+1)+lnx +\frac{2}{x}+C) =0\]

Answer 45

so, \[ y = \frac{-2\pm\sqrt {2^2-4*1*(the~~whole~bracket)}}{2}\]

Answer 46

got it?

Answer 47

it's a lot to take in

Answer 48

yes, but we are on the right track, right? lalalaala
from these, plug x =1 and y =1 into them to have C

Answer 49

It is the beginning of the semester, right?

Answer 50

sorry!

Answer 51

because there is a lot for you to delete.

Answer 52

yes

Answer 53

jamal is a boy name in america

Answer 54

con trai

Answer 55

not yet :( school first

Answer 56

that's good, son

Answer 57

tại sao tên của dì loser66?

Answer 58

How do you understand the word of "Loser"?? 66 is my born year

Answer 59

cảm ơn rất nhiều! nghe có vẻ rất tốt đẹp

Answer 60

Please, speak English!! Your Vietnamese is ridiculous as my English. hehehe

Answer 61

you said that sentence very well! hahaha

Answer 62

And don't forget correct my English, please

Answer 63

including vocabulary.

Answer 64

And don't forget to correct my English, please

Answer 65

It's getting late now and I have class tomorrow morning.
Thank you for helping me with math, it was nice getting you to know you!
I will see you soon when I have new questions, which I will. Good night!

Answer 66

you can send it from here

Answer 67

ok

Answer 68

Thank you very much!