Question

Please help solve the Initial Value Problem:
x^2(dy/dx)=(4x^2-x-2)/((x+1)(y+1) , y(1)=1

Answer 1

You need some sort of boundary condition, e.g

\(y(0) = \text{something}\)

Answer 2

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

Answer 3

okay so i've integrated both sides and no i'm kind of stuck

Answer 4

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

Answer 5

This is a non-linear equation. Are you sure you've written it right?

Answer 6

yes

Answer 7

its the equation i posted above

Answer 8

tears as i stare at the screen

Answer 9

Ok, you've done the intergal correctly.

You can get \(c\) by plugging in \(y=1\) and \(x=1\)

Then using this value of \(c\) you can actually factorise the \(y\) part and proceed t solve.

Answer 10

did you work this out on paper?

Answer 11

Iterative typing in expressions into wolfram alpha. I did it step by step though.

Answer 12

what is the value of c?

Answer 13

\[-\frac12 - \ln(8)\] if it's on the side with all the \(x\)'s

Answer 14

3/2=3ln2+ln1+2+c

Answer 15

\[c = 2-\frac32 - 3\ln(2) -\ln(1)\]
\[3\ln(2) = \ln(2^3) = \ln(8)\]
\[\ln(1) = 0\]

Answer 16

oops, I meant \(\frac32 - 2\) at the start there

Answer 17

so c=-1/2-ln8, do we just leave it as it is?

Answer 18

\[\frac2x+\ln(x)+3\ln(1+x) - \ln(8) = \frac12 (y^2 + 2y + 1)\]

This should now be solveable

Answer 19

im sorry where did the +1 come from on the right sight and why did you multiply by 1/2

Answer 20

Ok, so we got as far as
\[\frac2x+\ln(x)+3\ln(x+1)-\ln(8)-\frac12 = \frac12 y^2 + y\]
you can move the \(-\frac12\) over to the other side
\[\frac2x+\ln(x)+3\ln(x+1)-\ln(8) = \frac12 y^2 + y + \frac12\]
factorise out the \(\frac12\)
\[\frac2x+\ln(x)+3\ln(x+1)-\ln(8) = \frac12 (y^2 + 2y+ 1)\]
one step further: you can factorise the expression in \(y\)
\[\frac2x+\ln(x)+3\ln(x+1)-\ln(8) = \frac12 (y+1)^2\]

Answer 21

bless YOUR SOUL!

Answer 22

could you possibly finish the problem?

Answer 23

my brain is going to explode

Answer 24

but I COMPLETELY UNDERSTOOD what you just wrote

Answer 25

I'll tell you what to do:
multiply both sides by 2
square root both sides
subtract 1 from both sides
and you're done

Answer 26

and then plug in 1 for both x and y?

Answer 27

You've already done that bit to get \(c\)

Answer 28

i can SLEEP at night now!