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

Question

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

usukidoll · Answer

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

usukidoll · Answer

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

anonymous · Answer

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

anonymous · Answer

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 · Answer

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

usukidoll · Answer

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

anonymous · Answer

yes

usukidoll · Answer

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 · Answer

its only c^3

anonymous · Answer

1/8 c^3

usukidoll · Answer

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

usukidoll · Answer

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

usukidoll · Answer

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

anonymous · Answer

yeah

anonymous · Answer

sorry