Question-197129

Question Number 197129 by sonukgindia last updated on 08/Sep/23

Answered by Frix last updated on 09/Sep/23

This just comes from my experience with

trigonometric functions . I^{'} m not good at

solving differential equations .

f (x) = \tan x

\Rightarrow

f^{'} (x) = \frac{1}{\cos^{2} x} \land f^{″} (x) = \frac{2 \sin x}{\cos^{3} x}

\Rightarrow

\frac{f^{″} (x)}{f^{'} (x)} = 2 \tan x

Let y (x) = a \tan (b x + c)

\Rightarrow

y^{'} (x) = \frac{a b}{\cos^{2} (b x + c)} \land y^{″} (x) = \frac{2 {a b}^{2} \sin (b x + c)}{\cos^{3} (b x + c)}

\Rightarrow

\frac{f^{″} (x)}{f^{'} (x)} = 2 b \tan (b x + c)

Now find the constants \dots

Answered by qaz last updated on 09/Sep/23

y^{″} = 2 {y y}^{'} = {(y^{2})}^{'}

\Rightarrow y^{'} = y^{2} + C_{1} \Rightarrow \frac{1}{\sqrt{C_{1}}} \arctan \frac{y}{\sqrt{C_{1}}} = x + C_{2}

C_{1} = π C_{2} = \frac{\sqrt{π}}{4} \Rightarrow \frac{1}{\sqrt{π}} \arctan \frac{y}{\sqrt{π}} = x + \frac{\sqrt{π}}{4}

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