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WBJEE · Maths · Differential Equations

General solution of \(y \frac{d y}{d x}+b y^{2}=a \cos x\) \(0 \leq x < 1\) is (Here, \(c\) is an arbitrary constant)

  1. A \(y^{2}=2 a(2 b \sin x+\cos x)+c e^{-2 b x}\)
  2. B \(\left(4 b^{2}+1\right) y^{2}=2 a(\sin x+2 b \cos x)+c e^{-2 b x}\)
  3. C \(\left(4 b^{2}+1\right) y^{2}=2 a(\sin x+2 b \cos x)+c e^{2 b x}\)
  4. D \(y^{2}=2 a(2 b \sin x+\cos x)+c e^{-2 b x}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\left(4 b^{2}+1\right) y^{2}=2 a(\sin x+2 b \cos x)+c e^{-2 b x}\)

Step-by-step Solution

Detailed explanation

Given. \(y \frac{d y}{d x}+b y^{2}=a \cos x, 0 < x < 1\) \(y^{2}=z\) \(2 y \frac{d y}{d x}=\frac{d z}{d x}\) \(\Rightarrow \quad y \frac{d y}{d x}=\frac{1}{2} \frac{d z}{d x}\) \(\frac{1}{2} \frac{d z}{d x}+b y^{2}=a \cos x\) \(\frac{d z}{d x}+2 b y^{2}=2 a \cos x\) Now,…