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1. Find the solution to x^3 dy/dx = 2y^2

y=-4/3 and x=2

I have the answer which is (x^2)/(1-x^2) but i can't get to it.

2. (Original post by shle3brown)
Find the solution to x^3 dy/dx = 2y^2

y=-4/3 and x=2

I have the answer which is (x^2)/(1-x^2) but i can't get to it.

Begin by noticing that this is a separable equation, which can be rewritten as

Now integrate both sides and go from there. Post your working if you get stuck.
3. Try integrating it!

READ THIS IF YOU ARE STUCK AND WANT TO CHECK THE SOLUTION AND ANSWER!
Spoiler:
Show

x^3 dy/dx = 2y^2
(1/(x^3)) dx/dy = 1/(2y^2)
(1/(x^3))dx = (1/(2y^2))dy
∫(1/(x^3))dx = ∫(1/(2y^2))dy
Integrating both the sides will give you:
(-1/2(x^-2)) + c = -1/(2y)
(2x^2 - c) / 2 = y

Put in the values:
(2(2)^2 - c) /2 = -4/3
c = (8/3) + 8
c= 32/3

Put the value of c in the equation:
x^2 + 16/3 = y
in it's simplest form.

There might be some mistakes though, I apologize for that.

4. separate the variables as RDK as shown. if you prefer you could move the 2 to the left; it would appear as 1/2.

once the separation is complete you integrate both sides. you use a single constant of integration; this is usually written on the right hand side.
5. (Original post by YasinZahra)
Try integrating it!

x^3 dy/dx = 2y^2

There might be some mistakes though, I apologize for that.
Please amend your post - full solutions [no I didn't check it] are NOT allowed in the Maths forum.
6. (Original post by Muttley79)
Please amend your post - full solutions [no I didn't check it] are NOT allowed in the Maths forum.
OMG really? Like is this rule made by the TSR team? Sorry I'm not really aware of the rules and stuff here because I'm new.
7. (Original post by YasinZahra)
OMG really? Like is this rule made by the TSR team? Sorry I'm not really aware of the rules and stuff here because I'm new.
Yes, it is a TSR rule. Thanks for amending your post - please just give a hiint and don't even put the solution in a spoiler

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