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प्रश्न
Solve the differential equation dy = cosx(2 – y cosecx) dx given that y = 2 when x = `pi/2`
उत्तर
The given differential equation is dy = cosx(2 – y cosecx) dx
⇒ `"dy"/"dx"` = cosx(2 – y cosec x)
⇒ `"dy"/"dx"` = 2cosx – ycosx . cosecx
⇒ `"dy"/"dx"` = 2cosx – ycotx
⇒ `"dy"/"dx" + y cot x` = 2cosx
Here, P = cotx and Q = 2cosx.
∴ Integrating factor I.F. = `"e"^(intPdx)`
= `"e"^(int cot xdx)`
= `"e"^(log sinx)`
= sin x
∴ Required solution is `y xx "I"."F" = int "Q" xx "I"."F". "d"x + "c"`
⇒ `y . sin x = int 2 cos x . sin x "d"x + "c"`
⇒ `y . sin x = int sin 2x "d"x + "c"`
⇒ `y . sin x = - 1/2 cos 2x + "c"`
Put x = `pi/2` and y = 2, we get
`2 sin pi/2 = - 1/2 cos pi + "c"`
⇒ 2(1) = `- 1/2 (-1) + "c"`
⇒ 2 = `1/2 + "c"`
⇒ c = `2 - 1/2 = 3/2`
∴ The equation is y sin x = `- 1/2 cos 2x + 3/2`.
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