अवकल समीकरण \(ydx = \left( {y - x} \right)dy\) का सामान्य हल ज्ञात कीजिए। 

  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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संकल्पना:

पहली कोटि के एक रैखिक समीकरण के मानक रूप को \(\frac{{dy}}{{dx}} + Py = Q\) द्वारा ज्ञात किया गया है, जहाँ P, Q, x का स्वेच्छ फलन है। 

 

रैखिक समीकरण के समाकलन कारक को निम्न द्वारा ज्ञात किया गया है\(I.F. = {e^{\smallint pdx}}\)

रैखिक समीकरण के हल को निम्न द्वारा ज्ञात किया गया है\(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\)

गणना:

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

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

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

यह \(\frac{{dx}}{{dy}} + Px = Q\) का रूप है। 

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

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

रैखिक समीकरण के हल को निम्न द्वारा ज्ञात किया गया है

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

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

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

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