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

Answer: (D)

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