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anonymous

  • 5 years ago

Solve the differential equation by separation of variables y ln x (dx/dy) = (y+1/x)^2

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  1. anonymous
    • 5 years ago
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    If this got posted twice sry my tablet works funny on this site

  2. anonymous
    • 5 years ago
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    Is it a case where you don't know how to separate it, or do the actual integration?

  3. anonymous
    • 5 years ago
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    \[y \log x \frac{dy}{dy}=\left( \frac{y+1}{x} \right)^2 = \frac{(y+1)^2}{x^2} \rightarrow \frac{y}{(y+1)^2}dy=\frac{dx}{x^2 \log x}\]

  4. anonymous
    • 5 years ago
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    The fraction in y can be decomposed:\[\frac{y}{(y+1)^2}=\frac{1}{y+1}-\frac{1}{(y+1)^2}\]

  5. anonymous
    • 5 years ago
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    Damn...I misread dx/dy as dy/dx...hang on...

  6. anonymous
    • 5 years ago
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    It's simpler:\[y \log x \frac{dx}{dy}=\frac{(y+1)^2}{x^2} \rightarrow x^2 \log x dx = \frac{(y+1)^2}{y}dy\]

  7. anonymous
    • 5 years ago
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    For the left-hand side, make a substitution, u = log(x), then\[x=e^u \]and also\[dx = e^u du\]The right-hand side becomes:\[\int\limits_{}{}x^2 \log x dx = \int\limits_{}{}(e^{u})^2u.e^udu=\int\limits_{}{}u e^{3u}du\]This can be solved using integration by parts a couple of times (selecting u as your 'u' and e^u as your 'dv') to give\[\frac{e^{3u}}{3}(u-\frac{1}{3})+c=\frac{x^3}{3}(\log x - \frac{1}{3})+c\]

  8. anonymous
    • 5 years ago
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    The right-hand side becomes:\[\int\limits_{}{}\frac{(y+1)^2}{y}dy=\int\limits_{}{}\frac{y^2+2y+1}{y}dy=\int\limits_{}{}y + 2 + \frac{1}{y}dt=\frac{y^2}{2}+2y+\log y + c\]The constants are not necessarily this same, even though I've used c twice. Your solution is then\[\frac{y^2}{2}+2y+\log y = \frac{x^3}{3}(\log x - \frac{1}{3})+c\]

  9. anonymous
    • 5 years ago
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    Leave it like this.

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