Question

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck


`\sqrt{2 - x^(2)} \cdot (dy)/(dx) = (1)/(2 - y)`

f(1) = 0. Express y in terms of x

Answer 1

`\sqrt{2 - x^(2)} \cdot (dy)/(dx) = (1)/(2 - y)`

`(dy)/(dx) = \frac{1}{(2 - y)\sqrt{2 - x^(2)}}`

Now, isolate the variables, so you can integrate.

`(2 - y)dy = \frac{dx}{\sqrt{2 - x^(2)}}`

`\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^(2)}}`

`2y - (y^(2))/(2) = sin^(-1)(x)/(√(2)) + (1)/(2)C`



`4y - y^(2) = 2sin^(-1)(x)/(√(2)) + C`

`y^(2) - 4y = -2sin^(-1)(x)/(√(2)) - C`

`(y - 2)^(2) - 4 = -2sin^(-1)(x)/(√(2)) - C`

`(y - 2)^(2) = 4 - 2sin^(-1)(x)/(√(2)) - C`



`y - 2 = \pm\sqrt{4 - 2sin^(-1)(x)/(√(2)) - C}`

`y = 2 \pm\sqrt{4 - 2sin^(-1)(x)/(√(2)) - C}`

At x = 1, y = 0.

`0 = 2 \pm\sqrt{4 - 2sin^(-1)(1)/(√(2)) - C}`

`-2 = \pm\sqrt{4 - 2sin^(-1)(1)/(√(2)) - C}`


`4 - 2sin^(-1)(1)/(√(2)) - C > 0`

`\therefore 2 = \sqrt{4 - 2sin^(-1)(1)/(√(2)) - C}`



`4 = 4 - 2sin^(-1)(1)/(√(2)) - C`

`0 = -2sin^(-1)(1)/(√(2)) - C`

`C = -2sin^(-1)(1)/(√(2)) = -2(\pi)/(4)`

`C = -(\pi)/(2)`


`\therefore y = 2 - \sqrt{4 + (\pi)/(2) - 2sin^(-1)(x)/(√(2))}`