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

$$\mathrm{Determine}\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\mathrm{and}\mathrm{e}\mathrm{such}\mathrm{that}$$$$\left(1\right)\underset{x\to 0}{\lim }\frac{\cos \mathrm{ax}+{\mathrm{bx}}^3+{\mathrm{cx}}^2+\mathrm{dx}+\mathrm{e}}{{\mathrm{x}}^4}=\frac{2}{3}$$$$\left(2\right)\mathrm{find}\mathrm{general}\mathrm{solution}:{3\mathrm{xy}}^2{\mathrm{y}}^{\prime }={4\mathrm{y}}^3-{\mathrm{x}}^3$$

Answered by bobhans last updated on 09/Sep/20

$$\left(\mathrm{♠}\right){3\mathrm{xy}}^2{\mathrm{y}}^{\prime }={4\mathrm{y}}^3-{\mathrm{x}}^3$$$$\mathrm{setting}\mathrm{y}=\mathrm{zx}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{z}+\mathrm{x}\frac{\mathrm{dz}}{\mathrm{dx}}$$$$\Rightarrow \left(3\mathrm{x}\right)\left({\mathrm{z}}^2{\mathrm{x}}^2\right)(\mathrm{z}+\mathrm{x}\frac{\mathrm{dz}}{\mathrm{dx}})={4\mathrm{z}}^3{\mathrm{x}}^3-{\mathrm{x}}^3$$$$\Rightarrow {3\mathrm{z}}^2(\mathrm{z}+\mathrm{x}\frac{\mathrm{dz}}{\mathrm{dx}})={4\mathrm{z}}^3-1$$$$\Rightarrow {3\mathrm{z}}^3+{3\mathrm{z}}^2\mathrm{x}\frac{\mathrm{dz}}{\mathrm{dx}}={4\mathrm{z}}^3-1$$$$\Rightarrow {3\mathrm{z}}^2\mathrm{x}\frac{\mathrm{dz}}{\mathrm{dx}}={\mathrm{z}}^3-1;\frac{{3\mathrm{z}}^2}{{\mathrm{z}}^3-1}\mathrm{dz}=\frac{\mathrm{dx}}{\mathrm{x}}$$$$\Rightarrow \ln \mid {\mathrm{z}}^3-1\mid =\ln \mid \mathrm{Cx}\mid$$$$\Rightarrow {\mathrm{z}}^3=\mathrm{Cx}+1;\frac{{\mathrm{y}}^3}{{\mathrm{x}}^3}=\mathrm{Cx}+1$$$${\mathrm{y}}^3={\mathrm{Cx}}^4+{\mathrm{x}}^3\mathrm{or}\mathrm{y}=\sqrt[3]{{\mathrm{Cx}}^4+{\mathrm{x}}^3}$$

Answered by bemath last updated on 09/Sep/20

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