Question Number 197129 by sonukgindia last updated on 08/Sep/23
Answered by Frix last updated on 09/Sep/23
$$\mathrm{This}\:\mathrm{just}\:\mathrm{comes}\:\mathrm{from}\:\mathrm{my}\:\mathrm{experience}\:\mathrm{with} \\ $$$$\mathrm{trigonometric}\:\mathrm{functions}.\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{good}\:\mathrm{at} \\ $$$$\mathrm{solving}\:\mathrm{differential}\:\mathrm{equations}. \\ $$$${f}\left({x}\right)=\mathrm{tan}\:{x} \\ $$$$\Rightarrow \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}}\wedge{f}''\left({x}\right)=\frac{\mathrm{2sin}\:{x}}{\mathrm{cos}^{\mathrm{3}} \:{x}} \\ $$$$\Rightarrow \\ $$$$\frac{{f}''\left({x}\right)}{{f}'\left({x}\right)}=\mathrm{2tan}\:{x} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Let}\:{y}\left({x}\right)={a}\mathrm{tan}\:\left({bx}+{c}\right) \\ $$$$\Rightarrow \\ $$$${y}'\left({x}\right)=\frac{{ab}}{\mathrm{cos}^{\mathrm{2}} \:\left({bx}+{c}\right)}\wedge{y}''\left({x}\right)=\frac{\mathrm{2}{ab}^{\mathrm{2}} \mathrm{sin}\:\left({bx}+{c}\right)}{\mathrm{cos}^{\mathrm{3}} \:\left({bx}+{c}\right)} \\ $$$$\Rightarrow \\ $$$$\frac{{f}''\left({x}\right)}{{f}'\left({x}\right)}=\mathrm{2}{b}\mathrm{tan}\:\left({bx}+{c}\right) \\ $$$$\mathrm{Now}\:\mathrm{find}\:\mathrm{the}\:\mathrm{constants}… \\ $$
Answered by qaz last updated on 09/Sep/23
$${y}''=\mathrm{2}{yy}'=\left({y}^{\mathrm{2}} \right)' \\ $$$$\Rightarrow{y}'={y}^{\mathrm{2}} +{C}_{\mathrm{1}} \:\:\:\:\Rightarrow\frac{\mathrm{1}}{\:\sqrt{{C}_{\mathrm{1}} }}\mathrm{arctan}\:\frac{{y}}{\:\sqrt{{C}_{\mathrm{1}} }}={x}+{C}_{\mathrm{2}} \\ $$$${C}_{\mathrm{1}} =\pi\:\:\:\:\:{C}_{\mathrm{2}} =\frac{\sqrt{\pi}}{\:\mathrm{4}}\:\:\:\:\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\pi}}\mathrm{arctan}\:\frac{{y}}{\:\sqrt{\pi}}={x}+\frac{\sqrt{\pi}}{\mathrm{4}} \\ $$