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The solution of differential equation d t d x = t d d x ( g x ) - t 2 g ( x ) is:

Options

  1. At = g x + c x
  2. Bt = g x x + c
  3. Ct = g x x + c
  4. Dt = g x + x + c

Correct answer

C. t = g x x + c

Step-by-step solution

d t d x - t g ′ ( x ) g ( x ) = - t 2 g ( x ) ⇒   - 1 t 2 d t d x + 1 t   g ′ ( x ) g ( x ) = 1 g x ....(i) Let z = 1 t   ⇒   - 1 t 2 d t d x = d z d x ∴ From equation (i), d z d x + g ′ ( x ) g ( x ) z = 1 g x On comparing with d z d x + P z = Q , we get, P = g ′ ( x ) g ( x ) ,   Q = 1 g x ∴ I F = e ∫ g ′ ( x ) g ( x ) d x = e log ⁡ [ g x ] = g ( x ) Thus, the complete solution is, z ⋅ g x = ∫ g x ⋅ 1 g x

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