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प्रश्न
Solve: `2(y + 3) - xy "dy"/"dx"` = 0, given that y(1) = – 2.
उत्तर
Given differential equation is `2(y + 3) - xy "dy"/"dx"` = 0
⇒ `xy "dy"/"dx"` = 2y + 6
⇒ `(y/(2y + 6))"d"y = "dx"/x`
⇒ `1/2 (y/(y + 3))"d"y = "dx"/x`
Integrating both sides, we get
⇒ `1/2 int "y"/(y + 3) "d"y = int "dx"/x`
⇒ `1/2 int (y - 3 - 3)/(y + 3) "d"y = int "dx"/x`
⇒ `1/2 int (1 - 3/(y + 3))"d"y = int "dx"/x`
⇒ `1/2 int "d"y - 3/2 int 1/(y + 3) "d"y = int "dx"/x`
⇒ `1/2 y - 3/2 log |y + 3| = log x + "c"`
Put x = 1, y = –2
⇒ `1/2 (-2) - 3/2 log|-2 + 3| = log(1) + "c"`
⇒ `-1 - 3/2 log(1) = log(1) + "c"`
⇒ – 1 – 0 = 0 + c ....[∵ log (1) = 0]
∴ c = –1
∴ Equation is `1/2 y - 3/2 log|y + 3| = log x - 1`
⇒ `y - 3 log |y + 3| = 2 log x - 2`
⇒ `y - 3 log|(y + 3)^3| = log x^2 - 2`
⇒ `log|(y + 3)^3| + log x^2 = y + 2`
⇒ `log|x^2 (y + 3)^3| = y + 2`
⇒ `x^2(y + 3)^3 = "e"^(y + 2)`
Hence, the required solution is `x^2(y + 3)^3 = "e"^(y + 2)`
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