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Question
Solve the differential equation cos(x +y) dy = dx hence find the particular solution for x = 0 and y = 0.
Solution
Cos ( x + y )dy = dx
∴ `dy/dx = 1/[ cos ( x + y )]`
Let x + y = t
∴ 1 + `dy/dx = dt/dx`
∴ `dt/dx - 1 = 1/[ cos t ]`
`dt/dx = 1/[cost] + 1`
`dt/dx = [ 1 + cost ]/cost`
∴ `cost/[ 1 + cost ]dt = dx`
Integrating both side.
∴ `int cost/[ 1 + cost ]dt = int dx`
∴ `int [ cost( 1 - cost )]/sin^2t dt = x + c`
∴ `int (cosect.cot t - cot^2 t) dt = x + c`
∴ `int ( cosec t.cot t - cosec^2 t + 1 )dt = x + c`
∴ - cosect + cot t + t = x + c
∴ ` [cos t]/[sin t] - 1/[sin t] + t = x + c`
- tan`[( x + y )/2]` + x + y = x + c
∴ -tan`[( x + y )/2]`+ y = c
Putting x = 0, y = 0
∴ -tan`[( 0 + 0 )/2]`+ 0 = c
∴ c = 0
∴ y = tan`[( x + y )/2]`
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