Question

Solution of the differential equation $\frac{dy}{dx}+2y=\cos x$ is:
(where $C$ is integration constant)

• A. $y=\frac{1}{5}[\cos x+2\sin x]+C{e}^{-2x}$
• B. $y=\frac{4}{5}[2\cos x-\sin x]+C{e}^{-2x}$
• C. $y=\frac{1}{5}[\cos x-\sin x]+C{e}^{-2x}$
• D. $y=\frac{1}{5}[2\cos x+\sin x]+C{e}^{-2x}$

Answer 1

The correct option is D $y=\frac{1}{5}[2\cos x+\sin x]+C{e}^{-2x}$

It is a linear differential equation of the form
$\frac{dy}{dx}+Py=Q\left(x\right)$
where $P=2$ and $Q=\cos x$
Then $\text{IF}=e^{\int Pdx}=e^{\int 2dx}=e^{2x}$
Hence, the general solution is $y\left(\text{IF}\right)=\int Q\left(\text{IF}\right)dx$
$\Rightarrow y\cdot e^{2x}=\int e^{2x}\cos xdx+C$
On integrating By parts we get:
$\Rightarrow y\cdot e^{2x}=\frac{e^{2x}}{5}[2\cos x+\sin x]+C$
$\Rightarrow y=\frac{1}{5}[2\cos x+\sin x]+C{e}^{-2x}$