Question

If the solution curve of the differential equation `(dy)/(dx) = (x + y - 2)/(x - y)` passes through the point (2, 1) and (k + 1, 2), k > 0, then ______.

Options

• `2tan^-1(1/k) = log_e(k^2 + 1)`

• `tan^-1(1/k) = log_e(k^2 + 1)`

• `2tan^-1(1/(k + 1)) = log_e(k^2 + 2k + 2)`

• `2tan^-1(1/k) = log_e((k^2 + 1)/k^2)`

Answer 1

If the solution curve of the differential equation `(dy)/(dx) = (x + y - 2)/(x - y)` passes through the point (2, 1) and (k + 1, 2), k > 0, then `underlinebb(2tan^-1(1/k) = log_e(k^2 + 1))`.

Explanation:

Given differential equations is

`(dy)/(dx) = (x + y - 2)/(x - y) = ((x - 1) + (y - 1))/((x - 1) - (y - 1))`

Put x – 1 = h, y – 1 = k

`(dy)/(dx) = (h + k)/(h - k); k = Vh (dk)/(dh) = V + h(dV)/(dh)`

`V + X(dV)/(dh) = (1 + V)/(1 - V)  h(dV)/(dh) = (V^2 + 1)/(1 - V)`

`int(1 - V)/(1 + V^2) dV = int (dh)/h; int(dV)/(1 + V^2) - 1/2 int(2VdV)/(1 + V^2) = int (dh)/h`

`tan^-1V - 1/2 ln(1 + V^2)` = ln h + c

`tan^-1(k/h) - 1/2 ln(1 + k^2/h^2)` = ln h + c

Replace h and k by (x – 1) and (y – 1) respectively

`tan^-1((y - 1)/(x - 1)) - 1/2 ln (1 + (y - 1)^2/(x - 1)^2)` = ln(x – 1) + c

Satisfy (2, 1).

`0 - 1/2 ln 1` = ln 1 + c

∴ c = 0

Satisfy (k + 1, 2)

Therefore, `tan^-1(1/k) - 1/2 ln (1 + 1/k^2)` = ln k

`2tan^-1(1/k) = ln ((1 + k^2)/k^2) + 2 ln k`

`2tan^-1(1/k)` = ln (1 + k²)