Question

Let the solution curve of the differential equation `x (dy)/(dx) - y = sqrt(y^2 + 16x^2)`, y(1) = 3 be y = y(x). Then y(2) is equal to ______.

Options

• 15

• 11

• 13

• 17

Answer 1

Let the solution curve of the differential equation `x (dy)/(dx) - y = sqrt(y^2 + 16x^2)`, y(1) = 3 be y = y(x). Then y(2) is equal to 15.

Explanation:

Given differential equation is `x(dy)/(dx) - y = sqrt(y^2 + 16x^2)`

⇒ `x(dy)/(dx) = y + sqrt(y^2 + 16x^2)`

⇒ `(dy)/(dx) = (y + sqrt(y^2 + 16x^2))/x` ...(1)

Let y = tx

⇒ `(dy)/(dx) = t + x(dt)/(dx)` ...(2)

From equation (1) and (2), we get

`t + x(dt)/(dx) = (y + sqrt(y^2 + 16x^2))/x`

⇒ `t + x(dt)/(dx) = (tx + sqrt(t^2x^2 + 16x^2))/x`

⇒ `t + x(dt)/(dx) = t + sqrt(t^2 + 16)`

⇒ `x(dt)/(dx) = sqrt(t^2 + 16)`

⇒ `(dt)/sqrt(t^2 + 16) = (dx)/x`

Integrating both the sides, we get

`int(dt)/sqrt(t^2 + 16) = int(dx)/x`

⇒ In `[t + sqrt(t^2 + 16)]` = In x + In c

⇒ `t + sqrt(t^2 + 16)` = xc

⇒ `y/x + sqrt((y/x)^2 + 16)` = xc

⇒ `y/x + sqrt((y^2 + 16x^2)/x` = xc

⇒ `y + sqrt(y^2 + 16x^2)` = x²c

Now, y(1) = 3

⇒ `3 + sqrt(9 + 16)` = c

⇒ c = 8

⇒ `y + sqrt(y^2 + 16x^2)` = 8x²

So, at x = 2,

`y + sqrt(y^2 + 64)` = 32

⇒ `sqrt(y^2 + 64)` = (32 – y)

⇒ y² + 64 = y² + (32)² – 64y

⇒ 64 = 960

⇒ y = 15

∴ y(2) = 15