Let y=y(x) be the solution of the differential equation
(d y)/(d x)=2 x(x+y)³-x(x+y)-1, y(0)=1.
Then, (1/√2+y(1/√2))² equals :

Solution

Maths sir · 4 steps

The right side keeps showing x+y, so put u = x + y and the equation separates neatly.

1. Substitute u. Let u = x + y, so du/dx = 1 + dy/dx. The -1 on the right cancels the 1, leaving du/dx = 2xu³ - xu = xu(2u² - 1).
2. Separate and split. Write it as du/u(2u²-1) = x dx. Partial fractions give 1/u(2u²-1) = -1/u + 2u/(2u²-1).
3. Integrate. This gives -ln u + 1/2 ln(2u²-1) = x²/2 + C. Multiply by 2 and combine the logs to get ln(2 - 1/u²) = x² + C'. At x = 0, u = 0 + 1 = 1, so ln 1 = 0 and C' = 0.

   2 - 1/u² = e^(x²) ⇒ u² = 1/(2 - e^(x²))

4. So. At x = 1/√2, x² = 1/2, so (1/√2 + y(1/√2))² = u² = 1/(2 - √e). That is option (d).

The idea to keep. When x + y (or ax + by) keeps repeating in dy/dx, substitute u = x + y. Then du/dx = 1 + dy/dx, and the equation usually separates.

Answer from the NTA final answer key. Solution by PrepWiser.