Question

If y = y(x) is the solution of the differential equation `(1 + e^(2x))(dy)/(dx) + 2(1 + y^2)e^x` = 0 and y(0) = 0, then `6(y^'(0) + (y(log_esqrt(3))))^2` is equal to ______.

पर्याय

• 2

• –2

• –4

• –1

Answer 1

If y = y(x) is the solution of the differential equation `(1 + e^(2x))(dy)/(dx) + 2(1 + y^2)e^x` = 0 and y(0) = 0, then `6(y^'(0) + (y(log_esqrt(3))))^2` is equal to –4.

Explanation:

 `(1 + e^(2x))(dy)/(dx) + 2(1 + y^2)e^x` = 0 and y(0) = 0

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

⇒ `int(dy)/(1 + y^2) = -2inte^x/(1 + e^(2x))dx`

⇒ tan^–1(y) = –2tan^–1e^x + C

∵ y(0) = 0

∴ tan^–1(0) = –2tan^–1(1) + C

⇒ C = `2(π/4) = π/2`

⇒ tan^–1y + 2tan^–1e^x = `π/2`  ...(i)

Now, `((dy)/(dx))_(x = 0) = (-(2(1 + y^2)e^x)/(1 + e^(2x)))_(x = 0)`

⇒ `((dy)/(dx))_(x = 0)` = –1

⇒ Put x = `log_esqrt(3)` in equation (i), we get

`tan^-1y + 2tan^-1e^(log_e sqrt(3)) = π/2`

⇒ `tan^-1y + 2tan^-1sqrt(3) = π/2`

⇒ `tan^-1y + (2π)/3 = π/2`

⇒ tan^–1y = `–π/6`

⇒ `y(log_e sqrt(3)) = - 1/sqrt(3)`

∴ `6[y^'(0) + y(log_esqrt(3))^2] = 6[-1 + 1/3]` = –4