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

Commented by sonukgindia last updated on 07/Sep/23

$$checkonce$$

Commented by mokys last updated on 07/Sep/23

$$y^{″}=2{yy}^{\prime }\to y^{\prime }=y^2+c_1$$$$$$$$y^{\prime }\left(0\right)=2\pi \to c_1=4{\pi }^2-2\pi$$$$$$$$y^{\prime }=y^2+(4{\pi }^2-2\pi )$$$$$$$$\frac{dy}{y^2+(4{\pi }^2-2\pi )}=dx$$$$$$$$\frac{1}{\sqrt{4{\pi }^2-2\pi }}{tan}^{-1}\left(\frac{y}{\sqrt{4{\pi }^2-2\pi }}\right)=x+c_2$$$$$$$$y\left(0\right)=\sqrt{\pi }$$$$$$$$c_2=\frac{1}{\sqrt{4{\pi }^2-2\pi }}{tan}^{-1}\left(\frac{1}{\sqrt{4\pi -2}}\right)$$$$$$$$\therefore {tan}^{-1}\left(\frac{y}{\sqrt{4{\pi }^2-2\pi }}\right)=x\sqrt{4{\pi }^2-2\pi }+{tan}^{-1}\left(\frac{1}{\sqrt{4\pi -2}}\right)$$

Commented by sonukgindia last updated on 07/Sep/23

$$pleasedochecksecondline$$

Commented by mokys last updated on 07/Sep/23

$$noproblems$$

Commented by MathematicalUser2357 last updated on 10/Sep/23

$$checktwoce$$

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