hscishard
Active Member
Therefore x = [log(2x)]/2 ??
The question that involved this
WTF is wrong with my example?
Therefore x = [log(2x)]/2 ??
But then how can you get a derivative of something in terms of itself? [2nd example]i don't think you can do that. You're integrating 1/2x in respect to y not x
for your example to be correct , you are integrating (1/2x) with respect to y, NOT x !
Therefore x = [log(2x)]/2 ??
The question that involved this
WTF is wrong with my example?
x^2, doesn't really answer my question
therefore y = x + C
LOL
not sure what you mean, but in the first example this is what you're doingBut then how can you get a derivative of something in terms of itself? [2nd example]
for the second examplex^2, doesn't really answer my question
Notice how the dy is on the bottom? You can move it to the RHS of the equation and move the 2x on the otherside to the LHS and chuck integration operators in front to get an integration.x^2, doesn't really answer my question
Okay. I get it.
So..
x = logy + C
And
Is the final answer?
the first part is correctOkay. I get it.
So..
x = logy + C
And
Is the final answer?
Shit. Now I'm confused.the second part is not the answer
you can only integrate if its 1/2y, not 1/2x
In this one, you can't find r= (blah blah blah), can you?
WTF is wrong with my example?
The dy/dx = y I just amde up myself.what is the actual question, i dont think anyone actually knows
Shit. Now I'm confused.
In this one, you can't find r= (blah blah blah), can you?
But it's respect to twhy not, they would give an initial condition to find the C, then rearrange, take cubed root
you can, you just have to rearrange the equation you have for tIn this one, you can't find r= (blah blah blah), can you?