Maharashtra State BoardHSC Arts 12th Board Exam
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Find the particular solution of the following differential equation: dydxyx2xdydx-3ycotx=sin2x, when yy(π2)=2 - Mathematics and Statistics

Sum

Find the particular solution of the following differential equation:

`"dy"/"dx" - 3"y" cot "x" = sin "2x"`, when `"y"(pi/2) = 2`

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Solution

`"dy"/"dx" - 3"y" cot "x" = sin "2x"`

∴ `"dy"/"dx" - (3  "cot x")"y" = sin "2x"`   ....(1)

This is the linear differential equation of the form

`"dx"/"dy" + "Px" = "Q"` where P = `- 3 cot "x"` and Q = sin 2x.

∴ I.F. = `"e"^(int "P dy") = "e"^(int - 3 cot "x" "dx")`

`= "e"^(- 3 log sin "x") = "e"^(log (sin "x")^-3)` 

`= (sin "x")^-3 = 1/(sin^3"x")`

∴ the solution of (1) is given by

`"x" * ("I.F.") = int "Q" * ("I.F.") "dy" + "c"`

∴ `"y" xx 1/(sin^3 "x") = int sin "2x" xx 1/(sin "3x") "dx" + "c"`

∴ y cosec3 x = `int 2 sin "x" cos "x" xx 1/sin^3"x" "dx" + "c"`

∴ y cosec3 x = 2 `int (cos "x")/(sin^2 "x") "dx" + "c"`

Put sin x = t          ∴ cos x dx = dt

∴ y cosec3 x = 2`int 1/"t"^2 "dt" + "c"`

∴ y cosec3 x = 2`int "t"^-2 "dt" + "c"`

∴ y cosec3 x = 2`["t"^-1/-1] + "c"` 

∴ y cosec3 x = `(-2)/sin "x" + "c"`

∴ y cosec3 x + 2 cosec x = c

This is the general solution.

Now, `"y"(pi/2) = 2`, i.e. y = 2, when x = `pi/2`

∴ `2 "cosec"^3 pi/2 + 2 "cosec" pi/2 = "c"`

∴ 2(1)3 + 2(1) = c    

∴ c = 4

∴ the particular solution is

y cosec3 x + 2 cosec x = 4

∴ y cosec2 x + 2 = 4 sin x 

Notes

[Note: Answer in the textbook is incorrect.]

Concept: Formation of Differential Equations
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 6 Differential Equations
Miscellaneous exercise 6 | Q 6.3 | Page 218
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