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anonymous
 one year ago
ODE
y^2=cx+1/8 c^3  y=2xy'+y^2(y')^3
anonymous
 one year ago
ODE y^2=cx+1/8 c^3  y=2xy'+y^2(y')^3

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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3??? \[y^2=cx+\frac{1}{8}c^3..............y=2xy'+y^2(y')^3\] or \[y^2=cx+\frac{1}{8c^3}..............y=2xy'+y^2(y')^3\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I don't know what's going on.. can you go into more detail?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y^2=cx+1/8 c^3 y=2xy'+y^2(y')^3\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y^2=cx+1/8 c^3 y=2xy'+y^2(y')^3 if the given left is a solution od the ODE right :))

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so we are differentiating? one part of the equation with the fraction is still confusing.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so in the end we should have something like \[y=2xy'+y^2(y')^3 \]?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3could 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

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3_! \[y^2=cx+\frac{1}{8}c^3. \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3omg I apologize the latex is crazy dw:1436693748089:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so we let the x and y be the variables and we treat c as a constant.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3well 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
 one year ago
Best ResponseYou've already chosen the best response.0i dont know how to solve it :(

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3do you know the derivative of y^2 ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3hmm... I would just leave it as 2y unless we are using implicit differentiation as well?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3what about the derivative of cx?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3let'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) ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3well .. 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

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3why 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

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3uh ._. I have no idea what's going on now *mind blown*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01/8c^3 is not a constant

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.31/8 is a number... numbers are constants... derivatives of any number go to 0.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is the next step

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3hmmm is c also treated as a variable? I can't even...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3usually x and y are variables and anything but those letters are treated as a constant...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0uh ._. I have no idea what's going on now *mind blown* HAHA

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3in the end we are supposed to have x's and y's somehow.. so that c has to be treated as a constant.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0pass muna nahihirapan na ako e ahahaha

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3like... maybe factor one c out to make it like \[y^2 = c(x+\frac{1}{8}c^2) \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3then differentiate...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3like we know that the derivative of \[y^2 \rightarrow 2y y'\] and then product rule for c()

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is mean of ODE ??

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3ODE means an ordinary differential equation...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[y^2 dy = c(x+\frac{1}{8}c^2) dx\] if it's a separable equation, then we have to integrate both sides.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is the next step

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3integrate both sides.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh!! so we are given \[y=2xy' +y^{2}(y')^{3} and y ^{2}=cx+1\div8c ^{3} is \it ???\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3@sohailiftikhar supposedly we are differentiating \[y^2=cx+\frac{1}{8}c^3.\] and by doing that we get to \[y=2xy'+y^2(y')^3 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just put value of \[y ^{2}\] in equation of y then take derivative of it ...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3this isn't even my question, so .. yeah

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3and 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think we are give two equations and we have to find dy/dx

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3this is similar to OP's previous question though.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can't get it what is that .....sorry .

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I don't get it either.. let's get out of here.
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