Question

The solution of differential equation $\frac{dy}{dx}+3y={\cos }^{2}x$ is:

• A. $y=\frac{1}{6}+\frac{1}{26}(2\sin 2x+3\cos 2x)+c{e}^{-3x}$
• B. $y=\frac{1}{6}+\frac{1}{26}(2\sin 2x+3\cos 2x)+c{e}^{3x}$
• C. $y=\frac{1}{6}+\frac{1}{26}(2\sin 2x-3\cos 2x)+c{e}^{3x}$
• D. None of these

Answer 1

The correct option is A $y=\frac{1}{6}+\frac{1}{26}(2\sin 2x+3\cos 2x)+c{e}^{-3x}$

$\frac{dy}{dx}+3y={\cos }^{2}x$
$I.F.=e^{\int 3dx}=e^{3x}$
$\therefore$ the solution is $y.e^{3x}=\frac{1}{2}\int 2e^{3x}{\cos }^{2}xdx+c=\frac{1}{2}[e^{3x}(1+\cos 2x)+c]$
$=\frac{1}{2}[\frac{e^{3x}}{3}+\frac{e^{3x}}{9+4}][3\cos 2x+2\sin 2x]+c$
$\Rightarrow y=\frac{1}{6}+\frac{1}{26}(2\sin 2x+3\cos 2x)+c{e}^{-3x}$