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-

Question Number 112650 by bemath last updated on 09/Sep/20

Determine a, b, c, d and e such that

(1) \lim_{x \to 0} \frac{\cos ax + {bx}^{3} + {cx}^{2} + dx + e}{x^{4}} = \frac{2}{3}

(2) find general solution : {3 xy}^{2} y^{'} = {4 y}^{3} - x^{3}

Answered by bobhans last updated on 09/Sep/20

(♠) {3 xy}^{2} y^{'} = {4 y}^{3} - x^{3}

setting y = zx \Rightarrow \frac{dy}{dx} = z + x \frac{dz}{dx}

\Rightarrow (3 x) (z^{2} x^{2}) (z + x \frac{dz}{dx}) = {4 z}^{3} x^{3} - x^{3}

\Rightarrow {3 z}^{2} (z + x \frac{dz}{dx}) = {4 z}^{3} - 1

\Rightarrow {3 z}^{3} + {3 z}^{2} x \frac{dz}{dx} = {4 z}^{3} - 1

\Rightarrow {3 z}^{2} x \frac{dz}{dx} = z^{3} - 1; \frac{{3 z}^{2}}{z^{3} - 1} dz = \frac{dx}{x}

\Rightarrow \ln ∣ z^{3} - 1 ∣ = \ln ∣ Cx ∣

\Rightarrow z^{3} = Cx + 1; \frac{y^{3}}{x^{3}} = Cx + 1

y^{3} = {Cx}^{4} + x^{3} or y = \sqrt[3]{{Cx}^{4} + x^{3}}

Answered by bemath last updated on 09/Sep/20

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