Question

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

• A. $y\, sec\, x = tan\, x + c$
• B. $y \,tan \,x = sec \,x + c$
• C. $tan\, x = (sec \,x + c)y$
• D. $sec \,x = (tan \,x + c)y$

Answer 1

Answer: (D)

$sec \,x = (tan \,x + c)y$

Answer 2

$cos\, x \,dy = y\left(sin \,x - y\right) \,dx$ $\frac{dy}{dx} = y\, tan\,x \,y^{2}\, sec \,x$ $\frac{1}{y^{2}} \frac{dy}{dx} - \frac{1}{y} \,tan\, x = -sec \,x$ Let $\frac{1}{y} = t$ $-\frac{1}{y^{2}}\frac{dy}{dx}= \frac{dt}{dx}$ $-\frac{dy}{dx}- t \,tan\, x = -sec \,x \Rightarrow \frac{dt}{dx} + \left(tan\, x\right) \,t = sec\, x.$ $I.F. = e^{\int \,tan \,x\, dx} = sec\, x$ Solution is $t\left(I.F\right) = \int \left(I.F\right)\, sec\, x \,dx$ $\frac{1}{y} sec \,x = tan \,x + c$