Question

The solution of the differential equation $\sin y\frac{dy}{dx}=\cos y(1-x\cos y)$ is:

• A. $\sec y=x+1+K{e}^{-x}$
• B. $\sec y=x+1+K{e}^x$
• C. $\sec y=x-1+K{e}^x$
• D. $\sec y=x-1+K{e}^{-x}$

Answer 1

The correct option is B $\sec y=x+1+K{e}^x$

The given equation can be rewritten as
$\sec y\tan y\frac{dy}{dx}-\sec y=-x$. (dividing throughout by $co{s}^{2}y$).
Put $z=\sec$ $y$
$\therefore \frac{dz}{dx}=\sec y\tan y\frac{dy}{dx}$
Hence $\frac{dz}{dx}-z=-x$ which ls linear in $z$.
Hence the solution is $z{e}^{\int (-1)dx}=\int e^{\int (-1)dx}.(-x)dx$

$\Rightarrow z{e}^{-x}=-\int x{e}^{-x}dx=x{e}^{-x}-\int e^{-x}dx=x{e}^{-x}+e^{-x}+K$
$\therefore \sec y=x+1+K{e}^x$