By signing in you confirm you are 18 or older, or have your parent’s or guardian’s consent.

NDA 2 2025 Maths, question 94

Topic: Differential equations

Q94

If kk is an arbitrary constant, then what is the general solution of the equation (x+y)2dydx=k2(x+y)^{2}\frac{dy}{dx}=k^{2}?

  1. (a) y+x=tan⁡(x+c)+ky+x=\tan (x+c)+k
  2. (b) x+y=ktan⁡(y−ck)x+y=k\tan \left( \frac{y-c}{k} \right)Correct answer
  3. (c) x−y=ktan⁡(y−ck)x-y=k\tan \left( \frac{y-c}{k} \right)
  4. (d) y−x=tan⁡(x+c)+ky-x=\tan (x+c)+k

Solution.

Answer: (b) x+y=ktan⁡(y−ck)x+y=k\tan \left( \frac{y-c}{k} \right)

  1. Put u=x+yu = x + y and treat xx as a function of yy: dxdy=u2k2\frac{dx}{dy} = \frac{u^{2}}{k^{2}}, so dudy=1+u2k2\frac{du}{dy} = 1 + \frac{u^{2}}{k^{2}}.
  2. Separating and integrating: ktan⁡−1uk=y−ck\tan^{-1}\frac{u}{k} = y - c.
  3. So x+y=ktan⁡(y−ck)x + y = k\tan\left(\frac{y - c}{k}\right).

Trap: This needs k≠0k \neq 0; with k=0k = 0 the formula is undefined, but the question means a nonzero constant.