Question-197081

Question Number 197081 by sonukgindia last updated on 07/Sep/23

Commented by sonukgindia last updated on 07/Sep/23

$${check}\:{once} \\ $$

Commented by mokys last updated on 07/Sep/23

y′′ = 2yy′ → y′= y^2 +c_1 y′(0)= 2π → c_1 = 4 π^2 − 2π y′= y^2 + (4π^2 −2π) (dy/(y^2 +(4π^2 −2π))) = dx (1/( (√(4π^2 −2π)))) tan^(−1) ((y/( (√(4π^2 −2π)))))= x + c_2 y(0)=(√π) c_2 = (1/( (√(4π^2 −2π)))) tan^(−1) ( (1/( (√(4π−2))))) ∴ tan^(−1) ((y/( (√(4π^2 −2π)))))= x(√(4π^2 −2π)) + tan^(−1) ((1/( (√(4π−2)))))

$${y}''\:=\:\:\mathrm{2}{yy}'\:\rightarrow\:{y}'=\:{y}^{\mathrm{2}} +{c}_{\mathrm{1}} \\ $$$$ \\ $$$${y}'\left(\mathrm{0}\right)=\:\mathrm{2}\pi\:\rightarrow\:{c}_{\mathrm{1}} \:=\:\mathrm{4}\:\pi^{\mathrm{2}} \:−\:\mathrm{2}\pi \\ $$$$ \\ $$$$\:{y}'=\:{y}^{\mathrm{2}} \:+\:\left(\mathrm{4}\pi^{\mathrm{2}} −\mathrm{2}\pi\right) \\ $$$$ \\ $$$$\:\frac{{dy}}{{y}^{\mathrm{2}} +\left(\mathrm{4}\pi^{\mathrm{2}} −\mathrm{2}\pi\right)}\:=\:{dx} \\ $$$$ \\ $$$$\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}\pi^{\mathrm{2}} −\mathrm{2}\pi}}\:{tan}^{−\mathrm{1}} \left(\frac{{y}}{\:\sqrt{\mathrm{4}\pi^{\mathrm{2}} −\mathrm{2}\pi}}\right)=\:{x}\:+\:{c}_{\mathrm{2}} \: \\ $$$$ \\ $$$${y}\left(\mathrm{0}\right)=\sqrt{\pi} \\ $$$$ \\ $$$$\:{c}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}\pi^{\mathrm{2}} −\mathrm{2}\pi}}\:{tan}^{−\mathrm{1}} \left(\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}\pi−\mathrm{2}}}\right) \\ $$$$ \\ $$$$\therefore\:{tan}^{−\mathrm{1}} \left(\frac{{y}}{\:\sqrt{\mathrm{4}\pi^{\mathrm{2}} −\mathrm{2}\pi}}\right)=\:{x}\sqrt{\mathrm{4}\pi^{\mathrm{2}} −\mathrm{2}\pi}\:+\:{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}\pi−\mathrm{2}}}\right) \\ $$

Commented by sonukgindia last updated on 07/Sep/23