Question

Use separation of variables to solve the initial value problem. dy/dx = x/y and y =2 when x =1.

So far I've multiplied both sides by dx and got dy/y = x/y * y dx
then did the integral stuff and got
∫1/y dy = ∫x dx
which equals
∫y^-1 dy = ∫x dx
and that equals
-1/y^2 = 1

am I doing this right so far? and if i am, how do i continue this?

Answer 1

Nope. You need to multiply both sides by "dx" and both sides by "y" to get everything separated out.

ydy=xdx

That should be what you get from the algebra, right?

Answer 2

Oh okay. Was looking at my teachers example from this lesson and must have followed it wrong.
so after that I should plug the numbers and and get
2 = 1 + c
and that'd make C = 1, correct?

Answer 3

Well from this you're doing an integral... right?

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

Plugging in we get \[\frac{ 1 }{ 2 }2^2=\frac{ 1 }{ 2 }1^2 +C\] \[2=\frac{ 1 }{ 2 } +C\] \[\frac{ 3 }{ 2 } =C\]

Answer 4

So now you can just rewrite it as:

\[y^2=x^2+3\]

Answer 5

Ah, alright. Thank you. I got a final answer of \[\sqrt{x^{2}+3}\] and that appears to be correct according to my book. Thanks again!

Answer 6

Yeah, and remember when you take a square root you can have both positive and negative. So if your initial conditions were:

y=-2 and x=1 then you would do everything exactly the same except have:

\[y=-\sqrt{x^2+3}\]