Question

The general solution of the differential equation $\frac{dy}{dx}-y=\cos x$ is

• A. $y=\left(\frac{\sin x+\cos x}{2}\right)+C{e}^x$
• B. $y=\left(\frac{\sin x-\cos x}{2}\right)+C{e}^x$
• C. $y=\left(\frac{\sin x-\cos x}{2}\right)+C{e}^{-x}$
• D. $y=\left(\frac{\sin x+\cos x}{2}\right)+C{e}^{-x}$

Answer 1

Answer: (B)

Given differential equation is of the form
$\frac{dy}{dx}+Py=Q\text{ where }P=-1\text{ and }Q=\cos x$
Therefore, $IF=e^{\int -1dx}=e^{-x}$
On multiplying both sides of equation by IF, we get $e^{-x}\frac{dy}{dx}-e^{-x}y=e^{-x}\cos x$
or $\frac{dy}{dx}\left(y{e}^{-x}\right)=e^{-x}\cos x$
On integrating both sides w.r.t. $x$, we get
$y{e}^{-x}=\int e^{-x}\cos xdx+C.....$(i)
Let $I=\int e^{-x}\cos xdx$
$=\cos x\left(\frac{e^{-x}}{-1}\right)-\int (-\sin x)\left(-e^{-x}\right)dx$
$=-\cos x{e}^{-x}-\int \sin x{e}^{-x}dx$
$=-\cos x{e}^{-x}-\left[\sin x\left(-e^{-x}\right)-\int \cos x\left(-e^{-x}\right)dx\right]$
$=-\cos x{e}^{-x}+\sin x\cdot e^{-x}-\int \cos x\cdot e^{-x}dx$
$\Rightarrow I=-e^{-x}\cos x+\sin x{e}^{-x}-I$
$\Rightarrow 2I=(\sin x-\cos x)e^{-x}$
$\Rightarrow I=\frac{(\sin x-\cos x)e^{-x}}{2}$
Substituting the value of $I$ in Eq. (i), we get
$y{e}^{-x}=\left(\frac{\sin x-\cos x}{2}\right)e^{-x}+C$
$\Rightarrow y=\left(\frac{\sin x-\cos x}{2}\right)+C{e}^x$
which is the general solution of the given differential equation.