Question

The solution of differential equation $\frac{dy}{dx}-3y=sin\,2x$ is

• A. $y=e^{-3x}\left[\frac{cos\,2x+3\,sin\,2x}{13}\right]+c$
• B. $y=e^{-3x}\left(\frac{cos\,2x-3\,sin\,2x}{13}\right)+c$
• C. $ye^{-3x}=-e^{-3x} \frac{\left(2\,cos\,2x+3\,sin\,2x\right)}{13}+c$
• D. none of these

Answer 1

Answer: (C)

$ye^{-3x}=-e^{-3x} \frac{\left(2\,cos\,2x+3\,sin\,2x\right)}{13}+c$

Answer 2

$\frac{dy}{dx}-3y=sin\,2x$ $\Rightarrow$ It is linear equation with $I.F.=e^{\int-3dx}=e^{-3x}$ Required solution is, $y\cdot e^{-3x}=\int e^{-3x}\,sin\,2x\,dx$ $=e^{-3x} \frac{\left(-3\,sin\,2x-2\,cos\,2x\right)}{2^{2}+3^{2}}+c$ $\Rightarrow y\,e^{-3x}=-e^{-3x} \frac{\left(3\,sin\,2x+2\,cos\,2x\right)}{13}+c$