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

Answered by Frix last updated on 09/Sep/23

$$\mathrm{This}\mathrm{just}\mathrm{comes}\mathrm{from}\mathrm{my}\mathrm{experience}\mathrm{with}$$$$\mathrm{trigonometric}\mathrm{functions}.{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{good}\mathrm{at}$$$$\mathrm{solving}\mathrm{differential}\mathrm{equations}.$$$$f\left(x\right)=\tan x$$$$\Rightarrow$$$$f^{\prime }\left(x\right)=\frac{1}{{\cos }^{2}x}\wedge f^{″}\left(x\right)=\frac{2\sin x}{{\cos }^{3}x}$$$$\Rightarrow$$$$\frac{f^{″}\left(x\right)}{f^{\prime }\left(x\right)}=2\tan x$$$$$$$$$$$$\mathrm{Let}y\left(x\right)=a\tan (bx+c)$$$$\Rightarrow$$$$y^{\prime }\left(x\right)=\frac{ab}{{\cos }^2(bx+c)}\wedge y^{″}\left(x\right)=\frac{2{ab}^2\sin (bx+c)}{{\cos }^3(bx+c)}$$$$\Rightarrow$$$$\frac{f^{″}\left(x\right)}{f^{\prime }\left(x\right)}=2b\tan (bx+c)$$$$\mathrm{Now}\mathrm{find}\mathrm{the}\mathrm{constants}...$$

Answered by qaz last updated on 09/Sep/23

$$y^{″}=2{yy}^{\prime }={\left(y^2\right)}^{\prime }$$$$\Rightarrow y^{\prime }=y^2+C_1\Rightarrow \frac{1}{\sqrt{C_1}}\arctan \frac{y}{\sqrt{C_1}}=x+C_2$$$$C_1=\pi C_2=\frac{\sqrt{\pi }}{4}\Rightarrow \frac{1}{\sqrt{\pi }}\arctan \frac{y}{\sqrt{\pi }}=x+\frac{\sqrt{\pi }}{4}$$

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