Q. 215.0( 1 Vote )

# Find the particular solution of the following differential equation:

when x = 1

Answer :

The given differential equation is … (1)

Separating given differential equation, we get

On integrating, we get

⇒ y – 2 log (y + 2) = x + 2 log x + C ... (2)

Putting y = -1, when x = 1 in (2), we get

⇒ -1 – 2 log 1 = 1 + 2 log 1 + C

⇒ -1 – 0 = 1 + 0 + C

∴ C = -2

Equation (2) becomes y – 2 log (y + 2) = x + 2 log x – 2

⇒ y – x + 2 = 2 [log x + log (y + 2)]

⇒ y – x + 2 = 2 log {x (y + 2)}

∴ The required solution is y – x + 2 = 2 log {x (y + 2)}

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