Question

i just need a demonstration so steps (no answer) would be required

obtain the particular solution satisfying the initial condition indicated
\[2xyy' = 1 + y^2\] when \(x = 2, y = 3\)

Answer 1

\[2xyy'=1+y^2\]\[2xy\frac{dy}{dx}=1+y^2\]separate the variables.
That is to say treat the derivative in Leibniz notation as a fraction and split it up.
Get all the y's on one side and the x's on the other\[{ydy\over1+y^2}={2dx\over x}\]now integrate both sides...

Answer 2

sorry the 2 is on the wrong side\[{2ydy\over1+y^2}={dx\over x}\]the left side requires a u-sub
it makes no difference where you put the integration constant C; I'll put it on the RHS\[\ln

Answer 3

\[\ln (1 + x^2) = \ln x + c\] that right?

Answer 4

exactly :)
don't forget the integration constant!
this is where we finally use it

Answer 5

oh but that should have a y on the LHS
typo I'm sure

Answer 6

lol yeah =)) so \[\ln (\frac{1 + y^2}{x}) = c\] substitute right?

Answer 7

you could do it that way...

Answer 8

there's another?

Answer 9

but I wouldn't just in case x=0 is the initial condition ;)

Answer 10

exponentiate both sides

Answer 11

but x = 2 o.O

Answer 12

yes, but it may not always be

Answer 13

hmmm so then how do you exponentiate?

Answer 14

\[\ln (1 + y^2) = \ln x + C\]raise e to the power of each side to undo the logs

Answer 15

so \[1 + y^2 = e^{\ln x + C}??\]

Answer 16

yes, and don't forget that\[x^{a+b}=x^a\cdot x^b\]apply that to the RHS

Answer 17

\[1 + y^2 = x + e^c\]

Answer 18

careful...

Answer 19

\[1+y^2=e^c x\]

Answer 20

oh..oh yeah

Answer 21

but \(e^c\) is a constant, so we can just call it C again :P\[1+y^2=Cx\]

Answer 22

so \[e^c = \frac{1+y^2}{x}\] now substitute?

Answer 23

so \[C_1 = \frac{1+y^2}{x}\]

Answer 24

I recommend keeping x and y on separate sides; again if the initial condition was x=0 you'd be stuck here with that move

Answer 25

I'm trying to teach you a way that will always work, we could make exceptions here but that is a bad habit

Answer 26

but i need to get C to get particular solution right?

Answer 27

yes, but solving for c is usually quite easy by starting from the original equation\[1+y^2=Cx\implies 1+9=2C\implies C=5\]if x=0 we would have gotten 10=0, which would show no such C could exist

I suppose solving for C right away like you did suggests the same thing, but I think it is more clear the way I showed you
it's up to you

Answer 28

hmm well okay...since c = 5...what does that make the particular solution?

Answer 29

what was the general solution?
put a 5 where the C was
that's the particular solution

Answer 30

general solution is \[\ln (1+y^2) = \ln x + c\] right?

Answer 31

if you want it in that form you have to put the absolute value bars or you will change the domain

Answer 32

\[\ln |1 + y^2| = \ln |x| + c\] is the general solution?

Answer 33

yes, but it is equally valid exponentiated, so I would just write it as\[1+y^2=Cx\]

Answer 34

again, whatever floats your boat (or whatever your teacher prefers)

Answer 35

he doesnt prefer anything lol so basically i find the general solution..then from that solution i find C then substitute C into the general solution..that right?

Answer 36

exactly :)

Answer 37

let's say x = 2 and y = 3 are not given..what would the answer be then? would it just be the general solution?

Answer 38

yes, that would just leave us with the integration constant C undetermined, which, if you continue to study mathematics, corresponds to a "family" of function dependent on the possible values of C

Answer 39

...which is the general solution
so the answer to your Q is "yes"

Answer 40

good thanks :DD

Answer 41

welcome
you picked it up easily as usual
I'm out, later!