Question

Solution of the differential equation ${\left(\frac{dy}{dx}\right)}^2+\frac{dy}{dx}(cosx+secx)+1=0$ is/are:

• A. $y=-ln|secx+tanx|+c$
• B. $y=ln|secx+tanx|+c$
• C. $y=sinx+c$
• D. $y=cosx+c$

Answer 1

The correct option is A $y=-ln|secx+tanx|+c$

${\left(\frac{dy}{dx}\right)}^2+\frac{dy}{dx}(cosx+secx)+1=0$
$\Rightarrow \frac{dy}{dx}-\frac{-(cosx+secx)\pm \sqrt{(cosx-secx{)}^2}}{2}$
$\frac{dy}{dx}=\frac{-(cosx-secx)\pm (cosx-secx)}{2}$
$\frac{dy}{dx}=-secx$ & $\frac{dy}{dx}=-cosx$
$\Rightarrow y=-ln|secx+tanx|+c$ & $y=-sinx+c$