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zfleenor

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

  • one year ago
  • one year ago

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

    • one year ago
  2. zfleenor
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    The problem can be written in the form x + f(y) = C

    • one year ago
  3. heydayana
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    1/3 (y-4)^3 + C

    • one year ago
  4. heydayana
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    you're solving for the integral right?

    • one year ago
  5. zfleenor
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    i have to find f(y)

    • one year ago
  6. zfleenor
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    like in the form x + f(y) = C

    • one year ago
  7. ZeHanz
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    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.

    • one year ago
  8. ZeHanz
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    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.

    • one year ago
  9. jishan
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    good zeh.........

    • one year ago
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