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प्रश्न
Solve `dy/dx = (x+y+1)/(x+y-1) when x = 2/3 and y = 1/3`
उत्तर
`dy/dx = (x+y+1)/(x+y-1)` ...(i)
Put x + y = t ...(ii)
∴ y = t - x
Differentiating w.r.t.x, we get
∴ `dy/dx = dt/dx -1` ...(iii)
Substituting (ii) and (iii) in (i), we get
`dt/dx -1 = (t+1)/(t-1)`
∴ `dt/dx =(t+1)/(t-1) + 1 = (t+1+t-1)/(t-1)`
∴ `dt/dx = (2t)/(t-1)`
∴ `((t-1)/t)dt = 2dx`
∴ `(1-1/t)dt = 2dx`
Integrating on both sides, we get
`int (1-1/t) dt = 2 int dx`
∴ t - log | t | = 2x + c
∴ x + y -log |x + y| = 2x + c
x - 2x + y - log (x + y) = c
y - x - log (x + y) = c
y - x - c = log (x + y) ...(i)
Putting `x = 2/ 3 and y = 1/ 3` , we get
∴ from (i)
`1/3 - 2/3 - c = log (2/3 + 1/3)`
`(-1)/3 - c = log` (i)
`(-1)/3 - c = 0`
∴ c = `(-1)/3`
Put c = `1/ 3` in (i)
∴ c = `1/3`
`y - x - (-1/3)` = log (x + y)
`y - x +1/3` = log (x + y)
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