Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

zfleenor

  • 3 years ago

(dy/dx) = (y-4)^2

  • This Question is Open
  1. zfleenor
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[(dy/dx) = (y - 4)^{2}\]

  2. zfleenor
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. heydayana
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  4. heydayana
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you're solving for the integral right?

  5. zfleenor
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i have to find f(y)

  6. zfleenor
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  7. ZeHanz
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    good zeh.........

  10. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy