Question

use the separation of variables to solve the initial value problem:

dy/dx = x/y when y=2 and x = 1

Answer 1

no that is exactly what was given to me

Answer 2

That is a separable differential equation, so manipulate the equation such that all of the 'y' terms are on one side and all of the 'x' terms are on the other side:
Your equation:
\[\frac{dy}{dx} = \frac{x}{y}\]
\[\int{y \ dy} = \int{x \ dx}\]
It's an indefinite integral, so remember that there is a constant \(C\). You have to use your initial condition to solve for that constant.

Answer 3

oh, yeah, I see what they are asking you.

\(\large\color{black}{ \displaystyle \frac{dy}{dx} =\frac{x}{y} }\)

\(\large\color{black}{ \displaystyle y\frac{dy}{dx} =x }\)

\(\large\color{black}{ \displaystyle \color{blue}{\int\limits_{~}^{~}}y\frac{dy}{dx} \color{blue}{dx}=\color{blue}{\int\limits_{~}^{~}}x\color{blue}{dx} }\)

Answer 4

this is how it's done technically.

Answer 5

I integrated with respect to x on both sides, not dx's on the left side cancel

Answer 6

then integrate each side, and to find the +C plug in your values of y and x.

Answer 7

you can combine all your +C into one +C on the right side.

Answer 8

Okay I'm still a little sketchy on taking integral and the notation for that. I know it involves antiderivative but can we go through finding the integral?

Answer 9

Ok, what is \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}x~dx}\) ?

Answer 10

(just apply the power rule)

Answer 11

\[1/2 x^{2}\]

Answer 12

yes.

Answer 13

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}x~dx=\frac{1}{2}x^2+C}\)

Answer 14

And the left side, (after you cancel the dx's on both sides) is: \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}y~dx}\)

And, \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}y~dx=\frac{1}{2}y^2+C}\)

(integrating y with respect to y, and integrating x with respect to x -- is a same thing)

Answer 15

Okay add the constant. That makes sense. The rules for y are still confusing

Answer 16

for y it is just the same thing

Answer 17

okay so there is nothing special about the y at all? Alright then. So now you fill in x and y that were given to me and solve for c

Answer 18

yes, in this case there is nothing special about y.

Answer 19

(unlike, as you know adding a chain dy/dx for a derivative of y -- nothing like this when integrating)

Answer 20

Right that was what was confusing me on that part

Answer 21

and, then
\(\large\color{slate}{\displaystyle\frac{1}{2}y^2+C_1=\frac{1}{2}x^2+C_2}\)

\(\large\color{slate}{\displaystyle\frac{1}{2}y^2=\frac{1}{2}x^2+C_2-C_1}\)

\(\large\color{slate}{\displaystyle\frac{1}{2}y^2=\frac{1}{2}x^2+C_3}\)

C(3) -- still some arbitrary constant.

Answer 22

So your function just is:

\(\large\color{slate}{\displaystyle\frac{1}{2}y^2=\frac{1}{2}x^2+C}\)

Answer 23

you can divide by 1/2 everywhere

Answer 24

\(\large\color{slate}{\displaystyle y^2=x^2+C}\)

multiplying an arbitrary constant by 2 still leaves it an arbitrary constant

Answer 25

then:
\(\large\color{slate}{\displaystyle 2^2=1^2+C}\) >> \(\large\color{slate}{\displaystyle C=3}\)

Answer 26

hope this is making sense

Answer 27

Okay then my last question. Does it matter what form the function is in before plugging in? As in do I need to get the y by itself or anything or can I just start plugging in the variables I was given?

Answer 28

actually, I shouldn't multiply by 2 before finding C.

Answer 29

well, or I can by I should say:
\(\large\color{slate}{\displaystyle y^2=x^2+2C}\)

Answer 30

because this C that we got up to, is the actual vertical shift of the function.

Answer 31

so by multiplying times 2 everything instead of C I doubled the value of C (and I shouldn't do that)

Answer 32

\(\large\color{slate}{\displaystyle y^2=x^2+2C}\)
\(\large\color{slate}{\displaystyle 2^2=1^2+2C}\)
\(\large\color{slate}{\displaystyle 3=2C}\)
C=3/2

Answer 33

\(\large\color{slate}{\displaystyle y^2=x^2+3}\)

Answer 34

Alright I had to leave for a while. But I got it thank you SO MUCH!!!

Answer 35

surely, yw!