Question

Solution of the differential equation
$cosxdy=y(sinx-y)dx,0<x<\frac{\pi }{2}$ is

• A. $ysecx=tanx+c$
• B. $ytanx=secx+c$
• C. $tanx=(secx+c)y$
• D. $secx=(tanx+c)y$

Answer 1

The correct option is D $secx=(tanx+c)y$

$cosxdy=y(sinx-y)dx\frac{dy}{dx}=ytanx-y^{2}secx\frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}tanx=-secx$ ....(i)
Let $\frac{1}{y}=t\Rightarrow -\frac{1}{y^2}\frac{dy}{dx}=\frac{dt}{dx}$
From equations (i)
$-\frac{dt}{dx}-t\left(tanx\right)=-secx\Rightarrow \frac{dt}{dx}+\left(tanx\right)t=secx$
I.F.$=e^{\int tanxdx}=(e{)}^{log|secx|}secx$
Solution: $t(I.F)=\int (I.F)secxdx\Rightarrow \frac{1}{y}secx=tanx+c$