Question

`2("y" + 3) - x"y" "dy"/"dx"` = 0 को हल कीजिए जबकि y (1) = – 2 दिया है।

Answer 1

दिया गया अवकल समीकरण `2("y" + 3) - x"y" "dy"/"dx"` = 0

⇒ `x"y"  "dy"/"dx"` = 2y + 6

⇒ `("y"/(2"y" + 6)) "dy" = "dx"/x`

⇒ `1/2 ("y"/("y" + 3))"dy" = "dx"/x`

दोनों पक्षों को जोड़ने पर, हम प्राप्त करते हैं

⇒ `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"`

x = 1, y = –2 रखिए

⇒ `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

∴ समीकरण `1/2 "y" - 3/2 log|"y" + 3| = log x - 1` है।

⇒ `"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)`

इसलिए, वाँछित हल `x^2("y" + 3)^3 = "e"^("y" + 2)` है।