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Solve-x-2-1-dy-4x-xy-2-dx-y-0-2-




Question Number 9226 by tawakalitu last updated on 24/Nov/16
Solve: (x^2  + 1)dy = (4x + xy^2 )dx  y(0) = 2
$$\mathrm{Solve}:\:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\mathrm{dy}\:=\:\left(\mathrm{4x}\:+\:\mathrm{xy}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{2} \\ $$
Answered by mrW last updated on 24/Nov/16
(x^2 +1)dy=x(4+y^2 )dx  (1/(2^2 +y^2 ))dy=(x/(x^2 +1))dx  (1/2)tan^(−1) ((y/2))=(1/2)ln (x^2 +1)+(C/2)  tan^(−1) ((y/2))=ln (x^2 +1)+C  y(0)=2  tan^(−1) ((2/2))=ln (0+1)+C  C=(π/4)+nπ  tan^(−1) ((y/2))=ln (x^2 +1)+(π/4)+nπ  (y/2)=tan [ln (x^2 +1)+(π/4)+nπ]=tan [ln (x^2 +1)+(π/4)]  y=2∙tan [ln (x^2 +1)+(π/4)]
$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{dy}=\mathrm{x}\left(\mathrm{4}+\mathrm{y}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\mathrm{dy}=\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{y}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)+\frac{\mathrm{C}}{\mathrm{2}} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{y}}{\mathrm{2}}\right)=\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{C} \\ $$$$\mathrm{y}\left(\mathrm{0}\right)=\mathrm{2} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{2}}\right)=\mathrm{ln}\:\left(\mathrm{0}+\mathrm{1}\right)+\mathrm{C} \\ $$$$\mathrm{C}=\frac{\pi}{\mathrm{4}}+\mathrm{n}\pi \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{y}}{\mathrm{2}}\right)=\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)+\frac{\pi}{\mathrm{4}}+\mathrm{n}\pi \\ $$$$\frac{\mathrm{y}}{\mathrm{2}}=\mathrm{tan}\:\left[\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)+\frac{\pi}{\mathrm{4}}+\mathrm{n}\pi\right]=\mathrm{tan}\:\left[\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)+\frac{\pi}{\mathrm{4}}\right] \\ $$$$\mathrm{y}=\mathrm{2}\centerdot\mathrm{tan}\:\left[\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)+\frac{\pi}{\mathrm{4}}\right] \\ $$
Commented by tawakalitu last updated on 24/Nov/16
Thank you sir. God bless you.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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