A) \[sec\text{ }x=(tan\text{ }x+c)y\]
B) \[y\text{ }sec\text{ }x=tan\text{ }x+c\]
C) \[y\text{ }tan\text{ }x=sec\text{ }x+c\]
D) \[tan\text{ }x=(sec\text{ }x+c)y\]
Correct Answer: A
Solution :
\[\cos x\frac{dy}{dx}=y\sin x-{{y}^{2}}\] \[\cos x\frac{dy}{dx}-\sin x.y=-{{y}^{2}}\] \[\frac{1}{{{y}^{2}}}\frac{dy}{dx}-\frac{1}{y}\tan x=\sec x\] \[-\frac{1}{y}=z\] \[\frac{1}{{{y}^{2}}}\frac{dy}{dx}=\frac{dz}{dx}\] \[\frac{dz}{dx}+\tan x.z=\sec x\] \[I.F.={{e}^{\int{\tan x\,dx}}}\] Solution of above differential equation is \[z.\sec x=\int{{{\sec }^{2}}x\,dx}\] \[\frac{\sec x}{y}=\tan x+c\] \[\sec x=y(\tan x+c)\]
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