Question

Consider differential equation

Answer 1

\[\large y^2dx+(x-\dfrac{1}{y})dy=0\]

If y(1)=1 Then x=?

Answer 2

\[y^2dx+xdy-\frac1ydy=0\]

Answer 3

What I did is replace y bye 1 and took all x terms to one side and y to other. Then integrated. But I am not getting the answer.

Answer 4

I think the y(1)=1 is to be considered after solving for y for solving for the constant?

Answer 5

Okay so how do I proceed?

Answer 6

http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=y%5E2dx%2B(x-1%2Fy)dy%3D0

Answer 7

Doesn't exist (uses newly-defined function W(x))

Answer 8

Then how will I find x?

Answer 9

I mean the answer uses a newly-defined function, so it doesn't exist

Answer 10

Could you please double-check your question for any error?

Answer 11

Yes it is correct.

Answer 12

hmm... but wolfram can't be wrong...

Answer 13

its a first order linear equation, x(y) is the function

Answer 14

\(\large y^2dx+(x-\dfrac{1}{y})dy=0\)


\(\large \frac{dx}{dy}+\left(\frac{1}{y^2}\right)x = \dfrac{1}{y^3}=0\)

Answer 15

you can find IF and solve x

Answer 16

How do we understand that it is first order linear equation?

Answer 17

standard form of linear equaiton :

\((y ~terms )\times x' + (y~terms)\times x = (y~terms) \)

Answer 18

it looks like a standard form of linear equation in cartesian coordinates :

\(a x_1 + b x_2 = c\)

Answer 19

Thanks :)

Answer 20

initial condition must be x(1) = 1 right ?
i feel it cannot be y(1) = 1 ??

Answer 21

No. It is given y(1)=1

Answer 22

that must be a typo then, maybe double confirm wid whoever gave it agian...

IF method gives u a function "x(y)", but im not sure how u will be able to find the constant wid the info y(1) = 1

Answer 23

Then I shall take it as a typo.

Answer 24

solution is

\(x(y) = c~ e^{\frac{1}{y}} + \frac{1}{y} + 1\)

Answer 25

http://www.wolframalpha.com/input/?i=dx%2Fdy+%2B+x%2Fy%5E2+%3D+1%2Fy%5E3

Answer 26

lets think a bit before we conclude maybe...
a

Answer 27

\(x(y) = c~ e^{\frac{1}{y}} + \frac{1}{y} + 1 \)

\(y(1) = 1\)

we need to find \(c\)

Answer 28

Those are the options. C looks like the answer we are looking for.

Answer 29

ahh its a typo, x(1) = 1 is the initial condition
plugging (1, 1) gives c = -1/e... that matches wid option C

hope we arent missing anthing hmm...

Answer 30

we're not thinking actually :/
(1, 1) exists both on a function and its inverse...

so, if \(y(1) = 1 \iff y^{-1}(1) = x(1) = 1 \)

question is perfectly fine :)