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प्रश्न
Solve:
`2(y + 3) - xy (dy)/(dx)` = 0, given that y(1) = – 2.
Solve the differential equation `2(y + 3) - xy (dy)/(dx)` = 0, given y(1) = – 2.
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उत्तर
Given differential equation is `2(y + 3) - xy "dy"/"dx"` = 0
⇒ `xy (dy)/(dx)` = 2y + 6
⇒ `(y/(2y + 6))(dy) = (dx)/x`
⇒ `1/2 (y/(y + 3))(d)y = (dx)/x`
Integrating both sides, we get
⇒ `1/2 int y/(y + 3) dy = int dx/x`
⇒ `1/2 int (y - 3 - 3)/(y + 3) dy = int dx/x`
⇒ `1/2 int (1 - 3/(y + 3)) dy = int (dx)/x`
⇒ `1/2 int dy - 3/2 int 1/(y + 3) dy = 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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