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I don't know what's going on.. can you go into more detail?

\[y^2=cx+1/8 c^3 y=2xy'+y^2(y')^3\]

y^2=cx+1/8 c^3 y=2xy'+y^2(y')^3 if the given left is a solution od the ODE right :))

so we are differentiating? one part of the equation with the fraction is still confusing.

so in the end we should have something like \[y=2xy'+y^2(y')^3 \]?

yes

its only c^3

1/8 c^3

-_-! \[y^2=cx+\frac{1}{8}c^3. \]

omg I apologize the latex is crazy |dw:1436693748089:dw|

so we let the x and y be the variables and we treat c as a constant.

yeah

sorry

i dont know how to solve it :(

do you know the derivative of y^2 ?

2yy'

hmm... I would just leave it as 2y unless we are using implicit differentiation as well?

what about the derivative of cx?

c(1) >

c (1) + cx ?

\[2yy'= c+1/8 c^3\]

sorry

XD

\[3/8 c^2?\]

uh ._. I have no idea what's going on now *mind blown*

1/8c^3 is not a constant

1/8 is a number... numbers are constants... derivatives of any number go to 0.

what is the next step

hmmm is c also treated as a variable? I can't even...

usually x and y are variables and anything but those letters are treated as a constant...

3/8c^2 ?

uh ._. I have no idea what's going on now *mind blown*
HAHA

in the end we are supposed to have x's and y's somehow.. so that c has to be treated as a constant.

pass muna nahihirapan na ako e ahahaha

i dont understand

like... maybe factor one c out to make it like
\[y^2 = c(x+\frac{1}{8}c^2) \]

then differentiate...

like we know that the derivative of \[y^2 \rightarrow 2y y'\]
and then product rule for c()

what is mean of ODE ??

ODE means an ordinary differential equation...

what is the next step

integrate both sides.

oh!! so we are given \[y=2xy' +y^{2}(y')^{3} and y ^{2}=cx+1\div8c ^{3} is \it ???\]

just put value of \[y ^{2}\] in equation of y then take derivative of it ...

is that ??

this isn't even my question, so .. yeah

how it can be ...?

I think we are give two equations and we have to find dy/dx

this is similar to OP's previous question though.

I can't get it what is that .....sorry .

I don't get it either.. let's get out of here.

yes..