Question

The solution of the differential equation $\frac{dy}{dx} = (4x + y + 1)^{2}$ is

• A. 4x – y + 1 = 2 tan(2x – 2c)
• B. 4x – y – 1 = 2 tan(2x – 2c)
• C. 4x + y + 1 = 2 tan(2x + 2c)
• D. None of these

Answer 1

Answer: (C)

4x + y + 1 = 2 tan(2x + 2c)

Answer 2

Let 4x + y + 1 = z ⇒ $4 + \frac{dy}{dx} = \frac{dz}{dx}$ ⇒ $\frac{dy}{dx} = \frac{dz}{dx} - 4$

∴ $\frac{dy}{dx} = (4x + y + 1)^{2}$

⇒ $\frac{dz}{dx} - 4 = z^{2}$ ⇒ $\frac{dz}{dx} = z^{2} + 4$ ⇒ $\frac{dz}{z^{2} + 4} = dx$

⇒ $\frac{1}{2}\tan^{- 1}\frac{z}{2} = x + c$ ⇒ $\tan^{- 1}\left( \frac{4x + y + 1}{2} \right) = 2x + 2c$

∴ 4x + y + 1 = 2tan (2x + 2c)