d-2-y-dx-2-tan-x-dy-dx-0-

Question Number 175406 by princeDera last updated on 29/Aug/22

\frac{d^{2} y}{{d x}^{2}} - \tan (x) \frac{d y}{d x} = 0

Answered by Ar Brandon last updated on 29/Aug/22

y^{″} - \tan x \cdot y^{'} = 0

\Rightarrow \frac{y^{″}}{y^{'}} = \tan x

\Rightarrow \ln (y^{'}) = - \ln (\cos x) + C

\Rightarrow \frac{d y}{d x} = \frac{1}{\cos x} e^{C} = \frac{k}{\cos x}

\Rightarrow y = \int k \sec x d x = k \ln ∣ \sec x + \tan x) + k_{2}