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...
note:The tensors in the new coordinates system are represented by X'.
I have a question about the coordinate transformation of tensor.
σmn=amianjσ'ij; (1)
ωpq=apkaqlω'kl; (2)
In the original coordinates system, we have
σmn = Dmpnqωpq, (3)...