Answers

2015-05-19T22:01:39+05:30
3sinxy + 4cosxy = 5
⇒ 5(3/5 sinxy + 4/5 cosxy) = 5 
⇒ (3/5 sinxy + 4/5 cosxy) = 1
now (3/5)²+(4/5)² = 1
    so let, 3/5 =   cosA
             ⇒ 4/5 = sinA
So , (3/5 sinxy + 4/5 cosxy) = 1
     ⇒ (cosAsinxy + sinAcosxy) = 1
     ⇒ sin(A+xy) = 1
     ⇒ A + xy = 2πk + π/2 (k is any integer)
     ⇒ sin⁻¹(4/5) + xy = 2πk + π/2
     differenciating both sides with respect to x
   0 + xdy/dx + y = 0
      dy/dx = -y/x

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2015-05-20T20:42:58+05:30
3sinxy + 4cosxy = 5
⇒ 5(3/5 sinxy + 4/5 cosxy) = 5 
⇒ (3/5 sinxy + 4/5 cosxy) = 1
now (3/5)²+(4/5)² = 1
    so let, 3/5 =   cosA
             ⇒ 4/5 = sinA
So , (3/5 sinxy + 4/5 cosxy) = 1
     ⇒ (cosAsinxy + sinAcosxy) = 1
     ⇒ sin(A+xy) = 1
     ⇒ A + xy = 2πk + π/2 (k is any integer)
     ⇒ sin⁻¹(4/5) + xy = 2πk + π/2
     differenciating both sides with respect to x
   0 + xdy/dx + y = 0
      dy/dx = -y/x
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