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If `y^3-y=2x ,t h e n(x^2-1/(27))(d^2y)/(dx^2)+x(dy)/d=``y`b. `y/3`c. `y/9`d. `y/(27)`A. yB. `(y)/(3)`C. `(y)/(9)`D. `(y)/(27)` |
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Answer» Correct Answer - C We have `y^(3)-y=2x` Differentiating both sides w.r.t. x, we get `(3y^(2)-1)(dy)/(dx)=2 rArr (dy)/(dx)=(2)/(3y^(2)-1)" (1)"` Again, differentiating both sides w.r.t. x, we get `(d^(2)y)/(dx^(2))=(-2.6y((dy)/(dx)))/((3y^(2)-1))` Using (1), we get `(d^(2)y)/(dx^(2))=(-2y)/((3y^(2)-1)^(3))" (2)"` Now, `(x^(2)-(1)/(27))(d^(2)y)/(dx^(2))+(xdy)/(dx)` `=(x^(2)-(1)/(27))((-24y)/((3y^(2)-1)^(2)))+(2x)/((3y^(2)-1))` `=(y^(2)((y^(2)-1)^(2))/(4)-(1)/(27))((-24y)/((3y^(2)-1)^(3)))+(y(y^(2)-1))/(3y^(2)-1)` `" "(because y^(3)-y=2x)` `=({27y^(2)(y^(2)-1)^(2)-4})/(108)((-24y))/((3y^(2)-1))+(y(y^(2)-1))/(3y^(2)-1)` `=(y)/(9){(-54y^(2)(y^(2)-1)^(2)+8)/((3y^(2)-1)^(3))+(9(y^(2)-1))/(3y^(2)-1)}` `=(y)/(9){(-2(1+alpha)(alpha-2)^(2)+8)/(alpha^(3))}+(3(alpha-2))/(alpha)` `" where "alpha=3y^(2)-1` `=(y)/(9)` |
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