Here's the question you clicked on:
zfleenor
(dy/dx) = (y-4)^2
\[(dy/dx) = (y - 4)^{2}\]
The problem can be written in the form x + f(y) = C
you're solving for the integral right?
like in the form x + f(y) = C
Try separating variables:\[\frac{ dy }{ dx }=(y-4)^2 \rightarrow \frac{ dy }{ (y-4)^2 }=dx \]\[\int\limits_{}^{}\frac{ dy }{ (y-4)^2 }=\int\limits_{}^{}dx +C\]\[-\frac{ 1}{ y-4 }=x+C\]Solve for y:\[\frac{ 1 }{ y-4 }=C-x\]\[y-4=\frac{ 1 }{ C-x }\]\[y=\frac{ 1 }{ C-x }+4\]Where C is a real constant.
If you really want to rewrite it as \[x+f(y)=C\]in my calculation above, just before "solve for y" you could also write it as:\[x+-\frac{ 1 }{ y-4 }=C\]This means:\[f(y)=-\frac{ 1 }{ y-4 }\] In my view this is not an answer yet, because it is possible to get y as function of x, as you can see in the end of my calculation.