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Question-130301




Question Number 130301 by sarahvalencia last updated on 24/Jan/21
Answered by benjo_mathlover last updated on 24/Jan/21
(2) (dy/dx) = (x^2 /y^2 ) ⇒ ∫ y^2 dy−∫x^2 dx=C   y^3 −x^3  = 3C ; y^3 −x^3  = λ
$$\left(\mathrm{2}\right)\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{y}^{\mathrm{2}} }\:\Rightarrow\:\int\:\mathrm{y}^{\mathrm{2}} \mathrm{dy}−\int\mathrm{x}^{\mathrm{2}} \mathrm{dx}=\mathrm{C} \\ $$$$\:\mathrm{y}^{\mathrm{3}} −\mathrm{x}^{\mathrm{3}} \:=\:\mathrm{3C}\:;\:\mathrm{y}^{\mathrm{3}} −\mathrm{x}^{\mathrm{3}} \:=\:\lambda\: \\ $$
Commented by benjo_mathlover last updated on 24/Jan/21
(3) particular solution with y(1)=2   2−1 = λ ; λ=1  ∴ y^3  = x^3 +1
$$\left(\mathrm{3}\right)\:\mathrm{particular}\:\mathrm{solution}\:\mathrm{with}\:\mathrm{y}\left(\mathrm{1}\right)=\mathrm{2} \\ $$$$\:\mathrm{2}−\mathrm{1}\:=\:\lambda\:;\:\lambda=\mathrm{1} \\ $$$$\therefore\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{x}^{\mathrm{3}} +\mathrm{1}\: \\ $$

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