anonymous
  • anonymous
ODE y^2=cx+1/8 c^3 ------ y=2xy'+y^2(y')^3
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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UsukiDoll
  • 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
  • UsukiDoll
I don't know what's going on.. can you go into more detail?
anonymous
  • anonymous
\[y^2=cx+1/8 c^3 y=2xy'+y^2(y')^3\]

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anonymous
  • 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
  • UsukiDoll
so we are differentiating? one part of the equation with the fraction is still confusing.
UsukiDoll
  • UsukiDoll
so in the end we should have something like \[y=2xy'+y^2(y')^3 \]?
anonymous
  • anonymous
yes
UsukiDoll
  • 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
  • anonymous
its only c^3
anonymous
  • anonymous
1/8 c^3
UsukiDoll
  • UsukiDoll
-_-! \[y^2=cx+\frac{1}{8}c^3. \]
UsukiDoll
  • UsukiDoll
omg I apologize the latex is crazy |dw:1436693748089:dw|
UsukiDoll
  • UsukiDoll
so we let the x and y be the variables and we treat c as a constant.
anonymous
  • anonymous
yeah
anonymous
  • anonymous
sorry
anonymous
  • anonymous
@ganeshie8
UsukiDoll
  • 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
  • anonymous
i dont know how to solve it :(
UsukiDoll
  • UsukiDoll
do you know the derivative of y^2 ?
anonymous
  • anonymous
2yy'
UsukiDoll
  • UsukiDoll
hmm... I would just leave it as 2y unless we are using implicit differentiation as well?
UsukiDoll
  • UsukiDoll
what about the derivative of cx?
anonymous
  • anonymous
c(1) >
UsukiDoll
  • 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
  • anonymous
c (1) + cx ?
UsukiDoll
  • 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
  • anonymous
\[2yy'= c+1/8 c^3\]
UsukiDoll
  • 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
  • anonymous
sorry
anonymous
  • anonymous
XD
anonymous
  • anonymous
\[3/8 c^2?\]
UsukiDoll
  • UsukiDoll
uh ._. I have no idea what's going on now *mind blown*
anonymous
  • anonymous
1/8c^3 is not a constant
UsukiDoll
  • UsukiDoll
1/8 is a number... numbers are constants... derivatives of any number go to 0.
anonymous
  • anonymous
what is the next step
UsukiDoll
  • UsukiDoll
hmmm is c also treated as a variable? I can't even...
UsukiDoll
  • UsukiDoll
usually x and y are variables and anything but those letters are treated as a constant...
anonymous
  • anonymous
3/8c^2 ?
anonymous
  • anonymous
uh ._. I have no idea what's going on now *mind blown* HAHA
UsukiDoll
  • 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
  • anonymous
pass muna nahihirapan na ako e ahahaha
anonymous
  • anonymous
i dont understand
UsukiDoll
  • UsukiDoll
like... maybe factor one c out to make it like \[y^2 = c(x+\frac{1}{8}c^2) \]
UsukiDoll
  • UsukiDoll
then differentiate...
UsukiDoll
  • UsukiDoll
like we know that the derivative of \[y^2 \rightarrow 2y y'\] and then product rule for c()
sohailiftikhar
  • sohailiftikhar
what is mean of ODE ??
UsukiDoll
  • UsukiDoll
ODE means an ordinary differential equation...
UsukiDoll
  • 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
  • anonymous
what is the next step
UsukiDoll
  • UsukiDoll
integrate both sides.
sohailiftikhar
  • sohailiftikhar
oh!! so we are given \[y=2xy' +y^{2}(y')^{3} and y ^{2}=cx+1\div8c ^{3} is \it ???\]
UsukiDoll
  • 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
  • sohailiftikhar
just put value of \[y ^{2}\] in equation of y then take derivative of it ...
sohailiftikhar
  • sohailiftikhar
is that ??
UsukiDoll
  • UsukiDoll
this isn't even my question, so .. yeah
sohailiftikhar
  • sohailiftikhar
how it can be ...?
UsukiDoll
  • 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
  • sohailiftikhar
I think we are give two equations and we have to find dy/dx
UsukiDoll
  • UsukiDoll
this is similar to OP's previous question though.
sohailiftikhar
  • sohailiftikhar
I can't get it what is that .....sorry .
UsukiDoll
  • UsukiDoll
I don't get it either.. let's get out of here.
sohailiftikhar
  • sohailiftikhar
yes..

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