Recent content by Tensel

I find the problem. given that, \frac{dF(y)}{dy}=\frac{y^{2}}{y'} then, \frac{dF(x)}{dx}=\frac{dF(y)}{dy}.\frac{dy}{dx}=\frac{y^{2}}{y'}.y'=y^{2}, so, F(x)=\int y^{2}.dx=\int f^{2}(x).dx I made a mistake in the integration of F(x) i.e. F(x)≠\frac{f^{3}(x)}{3.f'(x)}+C But, what should be the...

pwsnafu, thank you for your reply. dF(y)/dy=y^2/y'-----this equation is what I want to construct, it can be considered as a known condition, I want to get F(y).

Yes, you are correct that K is not used, we can get rid of it. Yes, y' means the derivative of y with respect ot x. So, you meant that we should consider y' as constant when computing the derivative of F with respect to y? I need more information to understand this. At the beginning, we...

y=f(x),K=F(y), and dF(y)/dy=y^2/y', (1) then dF(x)=y^2*dx; so, F(x)=int(y^2*dx)=int((f(x)^2)*dx); then we obtain, F(x)=(f(x))^3/(3*f'(x))+C; substitution of y=f(x) into F(x), we get, F(y)=y^3/(3*y')+C; (2) using the result...

HallsofIvy, i want to calculate dF(y)/dy, not dF(y)/dx, but you remand me sth. thank you.

y =f(x), and y'=df(x)/dx, F(y)=y^3+(y')^2 how to deal with the (y')^2 when i calculate dF(y)/dy? thanks.