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Question Number 112650 by bemath last updated on 09/Sep/20
Determine a,b,c,d and e such that (1)lim_(x→0) ((cos ax+bx^3 +cx^2 +dx+e)/x^4 ) = (2/3) (2) find general solution : 3xy^2 y′ = 4y^3 −x^3
$$\mathrm{Determine}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\mathrm{and}\:\mathrm{e}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{ax}+\mathrm{bx}^{\mathrm{3}} +\mathrm{cx}^{\mathrm{2}} +\mathrm{dx}+\mathrm{e}}{\mathrm{x}^{\mathrm{4}} }\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{find}\:\mathrm{general}\:\mathrm{solution}\::\:\mathrm{3xy}^{\mathrm{2}} \:\mathrm{y}'\:=\:\mathrm{4y}^{\mathrm{3}} −\mathrm{x}^{\mathrm{3}} \: \\ $$
Answered by bobhans last updated on 09/Sep/20
(♠) 3xy^2 y′ = 4y^3 −x^3 setting y = zx ⇒(dy/dx) = z + x(dz/dx) ⇒(3x)(z^2 x^2 )(z+x (dz/dx)) = 4z^3 x^3 −x^3 ⇒3z^2 (z+x (dz/dx)) = 4z^3 −1 ⇒3z^3 + 3z^2 x (dz/dx) = 4z^3 −1 ⇒ 3z^2 x (dz/dx) = z^3 −1 ; ((3z^2 )/(z^3 −1)) dz = (dx/x) ⇒ ln ∣z^3 −1∣ = ln ∣Cx∣ ⇒ z^3 = Cx +1 ; (y^3 /x^3 ) = Cx +1 y^3 = Cx^4 +x^3 or y = ((Cx^4 +x^3 ))^(1/(3 ))
$$\left(\spadesuit\right)\:\mathrm{3xy}^{\mathrm{2}} \mathrm{y}'\:=\:\mathrm{4y}^{\mathrm{3}} −\mathrm{x}^{\mathrm{3}} \\ $$$$\:\:\mathrm{setting}\:\mathrm{y}\:=\:\mathrm{zx}\:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{z}\:+\:\mathrm{x}\frac{\mathrm{dz}}{\mathrm{dx}} \\ $$$$\Rightarrow\left(\mathrm{3x}\right)\left(\mathrm{z}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{z}+\mathrm{x}\:\frac{\mathrm{dz}}{\mathrm{dx}}\right)\:=\:\mathrm{4z}^{\mathrm{3}} \mathrm{x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{3z}^{\mathrm{2}} \left(\mathrm{z}+\mathrm{x}\:\frac{\mathrm{dz}}{\mathrm{dx}}\right)\:=\:\mathrm{4z}^{\mathrm{3}} −\mathrm{1}\: \\ $$$$\Rightarrow\mathrm{3z}^{\mathrm{3}} \:+\:\mathrm{3z}^{\mathrm{2}} \mathrm{x}\:\frac{\mathrm{dz}}{\mathrm{dx}}\:=\:\mathrm{4z}^{\mathrm{3}} −\mathrm{1}\: \\ $$$$\Rightarrow\:\mathrm{3z}^{\mathrm{2}} \mathrm{x}\:\frac{\mathrm{dz}}{\mathrm{dx}}\:=\:\mathrm{z}^{\mathrm{3}} −\mathrm{1}\:;\:\frac{\mathrm{3z}^{\mathrm{2}} }{\mathrm{z}^{\mathrm{3}} −\mathrm{1}}\:\mathrm{dz}\:=\:\frac{\mathrm{dx}}{\mathrm{x}} \\ $$$$\Rightarrow\:\mathrm{ln}\:\mid\mathrm{z}^{\mathrm{3}} −\mathrm{1}\mid\:=\:\mathrm{ln}\:\mid\mathrm{Cx}\mid\: \\ $$$$\Rightarrow\:\mathrm{z}^{\mathrm{3}} \:=\:\mathrm{Cx}\:+\mathrm{1}\:;\:\frac{\mathrm{y}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{3}} }\:=\:\mathrm{Cx}\:+\mathrm{1} \\ $$$$\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{Cx}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} \:\mathrm{or}\:\mathrm{y}\:=\:\sqrt[{\mathrm{3}\:}]{\mathrm{Cx}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} }\: \\ $$
Answered by bemath last updated on 09/Sep/20

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