পার্থক্যমূলক সমীকরণের সাধারণ সমাধান নির্ণয় করুন? \(ydx = \left( {y - x} \right)dy\)

  1. \(x = \frac{y}{2}\)
  2. \(x = \frac{y}{2} + \frac{c}{y}\)
  3. \(y = \frac{x}{2} + \frac{c}{x}\)
  4. \(y = \frac{x}{2}\)

Answer (Detailed Solution Below)

Option 2 : \(x = \frac{y}{2} + \frac{c}{y}\)
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Detailed Solution

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অনুসৃত ধারণা :

প্রথম ক্রমটির একটি রৈখিক সমীকরণের আদর্শ রূপ \(\frac{{dy}}{{dx}} + Py = Q\) দ্বারা প্রদত্ত যেখানে P, Q হল x এর নির্বিচারে কাজ।

রৈখিক সমীকরণের সমন্বিত উৎপাদক \(I.F. = {e^{\smallint pdx}}\) দ্বারা দেওয়া হয়

রৈখিক সমীকরণের সমাধান প্রদত্ত \(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\) দ্বারা।

গণনা :

\(ydx = \left( {y - x} \right)dy\)

\(y\frac{{dx}}{{dy}} = y - x\)

\(\frac{{dx}}{{dy}} + \frac{x}{y} = 1\)

এটি \(\frac{{dx}}{{dy}} + Px = Q\) এর রূপ।

\(I.F. = {e^{\smallint pdy}}\)

\(I.F. = {e^{lny}} = y\)

রৈখিক সমীকরণের সমাধান প্রদত্ত

\(x\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dy + c.\)

\(x\left( y \right) = \smallint 1\left( y \right)dy + c\)

\(xy = \frac{{{y^2}}}{2} + c\)

\(x = \frac{y}{2} + \frac{c}{y}\)
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