Question

solve this
dy/dx(1/y-logcosx)=-ytanx​

Answer 1

Answer:

$\log y - y\log\cos x = C$

Hello.

Hope this helps.

Step-by-step explanation:

The equation, in terms of differentials, is:

$\displaystyle\frac{dy}{y} - \log\cos x\, dy + y\tan x\,dx = 0$

There is a familiar product rule d(uv) = udv + vdu pattern in here.

Let u = - log cos x.

Then

$\displaystyle du = -d(\log\cos x) = -\frac{1}{\cos x}\times d(\cos x)\\ \\=-\frac{1}{\cos x}\times -\sin x\,dx = \tan x\,dx$

Thus the equation becomes

$\displaystyle\frac{dy}{y}+u\,dy + y\,du = 0\\ \\\Rightarrow \frac{dy}{y} + d(uy) = 0\\ \\\Rightarrow \log y + uy = C\\ \\\Rightarrow \log y - y\log\cos x = C$