Solve the First Order Differential Equation:$$\frac{\text dy}{\text dx}-y\tan x=-y^2\sec x$$

Question

unklerhaukus · Answer

$$\frac{\text dy}{\text dx}=y\tan x-y^2\sec x$$

anonymous · Answer

http://www.wolframalpha.com/input/?i=solve+dy%2Fdx%3Dytan(x)-y%5E2+sec(x)

unklerhaukus · Answer

$$\frac{\text dy}{\text dx}-y\tan x=-y^2\sec x$$
$$\text {let }y=\frac 1v$$
$$\frac{\text dy}{\text dx}=-\frac 1{v^2}\frac{\text dv}{\text dx}$$
$$-\frac 1{v^2}\frac{\text dv}{\text dx}-\frac{\tan x}v=-\frac{\sec x}{v^2}$$
$$\frac{\text dv}{\text dx}+v\tan x =\sec x$$
$$v^\prime+v\tan x=\sec x$$
$$R(x)=e^{\int \tan x\cdot\text dx}=e^{-\int\frac{-\sin x}{\cos x}\text dx}=e^{-\ln(\cos x)}=\sec x$$
$$\left(v\sec x\right)^\prime=\sec^2 x$$
$$v\sec x=\int \sec^2 x\text dx$$
$$v\sec x=\tan x+c$$
$$v=\sin x+c\cos x$$
$$\frac 1y=\sin x+c\cos x$$
$$y=\frac1{\sin x+c\cos x}$$