Question

The solution of the differential equation $\frac{dy}{dx}-2y\tan 2x=e^x\sec 2x$ is:

• A. y sin 2x = e${}^x$ + c
• B. y cos 2x = e${}^x$ + c
• C. y = e${}^x$ cos 2x + c
• D. y cos 2x + e${}^x$ = c

Answer 1

The correct option is B y cos 2x = e${}^x$ + c

The integrating factor is

$IF=e^{\int -2tan2xdx}$

$=e^{-lnsec2x}$

$=\frac{1}{sec2x}$

Hence multiplying the entire differential equation by IF gives us

$cos2x\frac{dy}{dx}-2ysin2x=e^x$

$cos2xdy-2ysin2xdx=e^{x}dx$

$d\left(ycos2x\right)=e^{x}dx$

$ycos2x=\int e^{x}dx$

$ycos2x=e^x+c$