dy/dx = x/y and y(3)=3, then find y.

Question

zepdrix · Answer

This is in your calculus class middle? :o
This is a separable differential equation, I guess you're getting an intro to these now huh :D

anonymous · Answer

Well, first that's a differential equation, type? separable variables, (i translate this in primitive, sorry for the bad english xD)
Now, multiply both sides by y

 y dy/dx = x
Then, integrate (both sides) on x

$$\int\limits y\dfrac {dy} {dx} dx=\int\limits xdx$$
Now, you'll obtain
$$\dfrac {y^2} {2}+c =\dfrac {x^2} {2} +c$$
Get together the constants, and

$$y= \pm \sqrt {2c+x^2}$$

Now, to find y you solve your IVP for y(3)=3, and remember that y is actually y(x)
So, you'll have that
$$3=\sqrt{2c+9}$$
And c must be 0
$$y=sqrt {x^2}$$

anonymous · Answer

@zepdrix haha yessss :(

anonymous · Answer

did he do it correctly?

zepdrix · Answer

Woops, combining the C's will give us a NEW arbitrary constant, not 2C.
 :D

And I think it's probably better to not square root this one. Just leave it implicitly written and plug in the initial condition to solve for C.

anonymous · Answer

but thank you @Umangiasd  :)

anonymous · Answer

Yeah, my mistake there, anyway, c=0 xD

zepdrix · Answer

$$\large \frac{1}{2}y^2=\frac{1}{2}x^2+c$$From here, multiply both sides by 2 giving us,$$\large y^2=x^2+c$$Understand why the constant didn't change?

anonymous · Answer

ok. uhm.. could you give me a few minutes? i am going to have to try this myself. :p

anonymous · Answer

so, when i did it, i got c=-5
and so, would the final answer be $$y=\sqrt{x^2-5}$$ ?

anonymous · Answer

@zepdrix

zepdrix · Answer

How did you get -5? :D

anonymous · Answer

i did |dw:1358117806034:dw|