Question

General solution of $y \frac{dy}{dx}+by^{2}=a\,cos\,x, 0 < x< 1$ is

• A. $y^{2} = 2a\left(2b\, sinx + cosx\right) + ce^{-2bx}$
• B. $\left(4b^{2} + 1\right)y^{2} = 2a\left(sinx + 2bcosx\right) + ce^{-2bx}$
• C. $\left(4b^{2} + 1\right)y^{2} = 2a\left(sinx + 2bcosx\right) + ce^{2bx}$
• D. $y^{2} = 2a\left(2bsinx + cosx\right) + ce^{2bx}$

Answer 1

Answer: (B)

$\left(4b^{2} + 1\right)y^{2} = 2a\left(sinx + 2bcosx\right) + ce^{-2bx}$

Answer 2

Let $y^{2} = z$ $y \frac{dy}{dx}=\frac{1}{2} \frac{dz}{dx}$ $\frac{dz}{dx}+2bz=2a\,cos\,x$ $IF=e^{2b\int dx}=e^{2bx}$ $z.e^{2bx}=\int\,2a\,cos\,x.e^{2bx}. dx$ $y^{2}e^{2bx}=\frac{2a}{4b^{2}+1}\left(sin\,x+2b\,cos\,x\right)e^{2bx}+c$ $\left(4b^{2}+1\right)y^{2}=2a\left(sinx+2bcox\right)+ce^{-2bx}$