## zfleenor 2 years ago (dy/dx) = (y-4)^2

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1. zfleenor

$(dy/dx) = (y - 4)^{2}$

2. zfleenor

The problem can be written in the form x + f(y) = C

3. heydayana

1/3 (y-4)^3 + C

4. heydayana

you're solving for the integral right?

5. zfleenor

i have to find f(y)

6. zfleenor

like in the form x + f(y) = C

7. ZeHanz

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.

8. ZeHanz

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.

9. jishan

good zeh.........