- #1
proto
- 5
- 0
[tex](2x^2ylny-x)y'=y[/tex]
[tex](2x^2ylny-x)dy=ydx[/tex]
then i divide both sides by dy
[tex](2x^2ylny-x)=yx'[/tex]
then i divide both sides by y
[tex](2x^2lny-\frac{x}{y})=x'[/tex]
[tex]x'+\frac{x}{y}=2x^2lny[/tex]
so i have here a bernuly foruma
i divide both sides by [tex]x^2[/tex]
[tex]\frac{x'}{x^2}+\frac{1}{xy}=2lny[/tex]
[tex]z=x^{-1}[/tex]
[tex]z'=-1x^{-2}x'[/tex]
[tex]-z'+\frac{z}{y}=2ln y[/tex]
z is defined to be a function of x
so [tex]z'=\frac{dz}{dx}[/tex]
why the book interprets [tex]z'=\frac{dz}{dy}[/tex]
??
z is linked to y not in a direct way .
but z linked to x in a direct way
z and x are more close to each other.
i can't see a mathematical way of figuring it out
its all intuition.and i my intuition is very bad
[tex](2x^2ylny-x)dy=ydx[/tex]
then i divide both sides by dy
[tex](2x^2ylny-x)=yx'[/tex]
then i divide both sides by y
[tex](2x^2lny-\frac{x}{y})=x'[/tex]
[tex]x'+\frac{x}{y}=2x^2lny[/tex]
so i have here a bernuly foruma
i divide both sides by [tex]x^2[/tex]
[tex]\frac{x'}{x^2}+\frac{1}{xy}=2lny[/tex]
[tex]z=x^{-1}[/tex]
[tex]z'=-1x^{-2}x'[/tex]
[tex]-z'+\frac{z}{y}=2ln y[/tex]
z is defined to be a function of x
so [tex]z'=\frac{dz}{dx}[/tex]
why the book interprets [tex]z'=\frac{dz}{dy}[/tex]
??
z is linked to y not in a direct way .
but z linked to x in a direct way
z and x are more close to each other.
i can't see a mathematical way of figuring it out
its all intuition.and i my intuition is very bad