A) tan x
B) sec x
C) cosec x
D) cot x
Correct Answer: B
Solution :
Consider the differential equation \[\frac{dy}{dx}=y\,tan\,x\,-{{y}^{2}}\,\sec \,x\] Divide by \[{{\operatorname{y}}^{2}}\] on both the sides, we get \[\frac{1}{{{y}^{2}}}\left( \frac{dy}{dx} \right)=\frac{\tan \,x}{y}-\sec \,\,x\] ... (1) Let \[\frac{1}{y}=z\] Differentiating both sides, we get: \[\frac{-1}{{{y}^{2}}}.\frac{dy}{dx}=\frac{dz}{dx}\] Put value of \[\frac{1}{{{y}^{2}}}\,\,\frac{dy}{dx}\] in the equation (1), we get \[-\left( \frac{dz}{dx} \right)-(tan\,x)z=-sec\,x\] \[\Rightarrow \,\,\,\left( \frac{dz}{dx} \right)+(tan\,x)z=\,\,sec\,x\] This is the linear diff equation in ?z? i.e. This is of the form \[\frac{dz}{dx} + P.z = Q\] \[\operatorname{then} integrating factor = {{e}^{\int{\,Pdx}}}\] \[\therefore \] In the given question \[I.F.={{e}^{\int{\tan x\,dx}}}={{e}^{\log (\sec \,x)}}=\sec \,\,x\]
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