हल कीजिए: \(\frac {dy}{dx} \cos (x - y) = 1\)

  1. \(\tan \left( \frac {x - y}{2} \right) = y + c\)
  2. \(\cot \left( \frac {y - x}{2} \right) = y + c\)
  3. -sin (x - y) = y + c
  4. \(\cot \left( \frac {x - y}{2} \right) = y + c\)

Answer (Detailed Solution Below)

Option 4 : \(\cot \left( \frac {x - y}{2} \right) = y + c\)
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संकल्पना:

दिए गए अवकल समीकरण का हल ज्ञात करने के लिए दिए गए अवकल समीकरण को चर वियोज्य रूप में परिवर्तित कीजिए और फिर इसका समाकलन कीजिए। 

गणना:

दिया गया है, \(\rm \frac {dy}{dx} \cos (x - y) = 1\)

⇒ \(\rm \dfrac {dx}{dy} = \cos (x - y)\)....(1)

 x - y = u रखने पर 

y के संबंध में अवकलन करने पर, हमें निम्न प्राप्त होता है 

⇒ \(\rm \dfrac {dx}{dy} - 1 = \dfrac {du}{dy}\)

⇒ \(\rm \dfrac {dx}{dy} = 1 + \dfrac {du}{dy}\)

समीकरण (1) निम्न हो जाता है,

\(\rm 1 + \dfrac {du}{dy} = cos \;u\)

\(\rm \dfrac {du}{dy} = cos \;u -1\)

\(\rm \dfrac {du}{cos \;u -1} = dy\)

चर अलग हैं। दोनों पक्षों का समाकलन करने पर, हमें निम्न प्राप्त होता है 

\(\rm \int \dfrac {du}{cos \;u -1} =\int dy\)

\(\rm \int \dfrac {du}{-2\;sin^2 \dfrac u 2} =\int dy\)

\(\rm -\dfrac 12 \int cosec^2 (\dfrac u 2)\; du =\int dy\)

\(\rm ⇒\cot \left( \frac {u}{2} \right) = y + c\)

\(\cot \left( \frac {x - y}{2} \right) = y + c\)

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