Question 1167011
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\frac{dy}{dx}\ =\ x(y^4\ +\ 2y^2\ +1)]


Initial condition: *[tex \LARGE y(-3)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y}{y^4\ +\ 2y^2\ +1}\,dy\ =\ x\,dx]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,\frac{y}{y^4\ +\ 2y^2\ +1}\,dy\ =\ \int\,x\,dx]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{1}{y^2\,+\,1}\ =\ \frac{x^2}{2}\ +\ c]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{\sqrt{-x^2\,-\,2c\,-1}}{\sqrt{x^2\,+\,2c}}\ \ ] or *[tex \LARGE y\ =\ \frac{\sqrt{-x^2\,-\,2c\,-1}}{\sqrt{x^2\,+\,2c}}\ \ ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(-3)\ =\ -\frac{\sqrt{-(-3)^2\,-\,2c\,-1}}{\sqrt{(-3)^2\,+\,2c}}\ =\ 1\ \ ] No solution for *[tex \LARGE c].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(-3)\ =\ \frac{\sqrt{-(-3)^2\,-\,2c\,-1}}{\sqrt{(-3)^2\,+\,2c}}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ =\ -\frac{19}{4}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \sqrt{\frac{17\,-\,2x^2}{2x^2\ -\ 19}}]




																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish]


I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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