ODE
y^2=cx+1/8 c^3 ------ y=2xy'+y^2(y')^3

- anonymous

ODE
y^2=cx+1/8 c^3 ------ y=2xy'+y^2(y')^3

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- UsukiDoll

??? \[y^2=cx+\frac{1}{8}c^3..............y=2xy'+y^2(y')^3\]
or
\[y^2=cx+\frac{1}{8c^3}..............y=2xy'+y^2(y')^3\]

- UsukiDoll

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

- anonymous

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

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## More answers

- anonymous

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

- UsukiDoll

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

- UsukiDoll

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

- anonymous

yes

- UsukiDoll

could you rewrite the first equation.. it's the fraction portion that is making me lost. I don't know if \[c^3y \] is in the denominator or the numerator

- anonymous

its only c^3

- anonymous

1/8 c^3

- UsukiDoll

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

- UsukiDoll

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

- UsukiDoll

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

- anonymous

yeah

- anonymous

sorry

- anonymous

@ganeshie8

- UsukiDoll

well on the farthest term the one with the 1/8c^3 if c is a constant and if all numbers are constants.. we will just have 0

- anonymous

i dont know how to solve it :(

- UsukiDoll

do you know the derivative of y^2 ?

- anonymous

2yy'

- UsukiDoll

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

- UsukiDoll

what about the derivative of cx?

- anonymous

c(1) >

- UsukiDoll

let's see... product rule is f(x)g'(x)+f'(x)g(x)
so we let f(x) = c and g(x) = x
what is the derivative of f'(x) and what is the derivative of g'(x) ?

- anonymous

c (1) + cx ?

- UsukiDoll

well .. the derivative of g(x) is 1 yes..
c(1) +
that part is correct
so for f'(x)g(x)
f(x) = c and g(x) = x
so we leave g(x) alone and take the derivative of f(x) = c. Since c is a constant, the derivative is 0 so we have
c(1) +x(0)
c+0
c

- anonymous

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

- UsukiDoll

why is 1/8c^3 still there? a number like 1/8 is just a constant and c^3 is a constant too? I'm lost as the hills XD

- anonymous

sorry

- anonymous

XD

- anonymous

\[3/8 c^2?\]

- UsukiDoll

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

- anonymous

1/8c^3 is not a constant

- UsukiDoll

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

- anonymous

what is the next step

- UsukiDoll

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

- UsukiDoll

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

- anonymous

3/8c^2 ?

- anonymous

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

- UsukiDoll

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

- anonymous

pass muna nahihirapan na ako e ahahaha

- anonymous

i dont understand

- UsukiDoll

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

- UsukiDoll

then differentiate...

- UsukiDoll

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

- sohailiftikhar

what is mean of ODE ??

- UsukiDoll

ODE means an ordinary differential equation...

- UsukiDoll

\[y^2 dy = c(x+\frac{1}{8}c^2) dx\] if it's a separable equation, then we have to integrate both sides.

- anonymous

what is the next step

- UsukiDoll

integrate both sides.

- sohailiftikhar

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

- UsukiDoll

@sohailiftikhar supposedly we are differentiating \[y^2=cx+\frac{1}{8}c^3.\] and by doing that we get to \[y=2xy'+y^2(y')^3 \]

- sohailiftikhar

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

- sohailiftikhar

is that ??

- UsukiDoll

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

- sohailiftikhar

how it can be ...?

- UsukiDoll

and with vague instructions, I can't go further... it's like hello I'm not a mind reader like what does the question want really?

- sohailiftikhar

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

- UsukiDoll

this is similar to OP's previous question though.

- sohailiftikhar

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

- UsukiDoll

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

- sohailiftikhar

yes..

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