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Question Number 108018 by Ar Brandon last updated on 13/Aug/20

i. x^3 (dy/dx)+y^2 +x^2 y+2x^4 =0 ii. (dy/dx)=−2−y+y^2 iii. 2cos(x)(dy/dx)=2cos^2 (x)−sin^2 (x)+y^2 ; y(0)=−1

$$\mathrm{i}.\:\:\mathrm{x}^{\mathrm{3}} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \mathrm{y}+\mathrm{2x}^{\mathrm{4}} =\mathrm{0} \\ $$$$\mathrm{ii}.\:\:\frac{\mathrm{dy}}{\mathrm{dx}}=−\mathrm{2}−\mathrm{y}+\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{iii}.\:\:\mathrm{2cos}\left(\mathrm{x}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{2cos}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{y}^{\mathrm{2}} \:;\:\mathrm{y}\left(\mathrm{0}\right)=−\mathrm{1} \\ $$

Answered by john santu last updated on 14/Aug/20

((♣JS♥)/∮) (ii) (dy/(y^2 −y−2)) = dx (dy/((y−2)(y+1))) = dx (((y+1)−(y−2))/(3(y−2)(y+1))) dy = dx ∫ (dy/(y−2))−∫(dy/(y+1)) = ∫ 3dx ln (((y−2)/(y+1))) = 3x +c ((y−2)/(y+1)) = Ce^(3x)

$$\:\:\:\:\:\frac{\clubsuit\mathcal{JS}\heartsuit}{\oint} \\ $$$$\left({ii}\right)\:\frac{{dy}}{{y}^{\mathrm{2}} −{y}−\mathrm{2}}\:=\:{dx} \\ $$$$\:\:\:\:\:\:\frac{{dy}}{\left({y}−\mathrm{2}\right)\left({y}+\mathrm{1}\right)}\:=\:{dx} \\ $$$$\:\:\:\:\frac{\left({y}+\mathrm{1}\right)−\left({y}−\mathrm{2}\right)}{\mathrm{3}\left({y}−\mathrm{2}\right)\left({y}+\mathrm{1}\right)}\:{dy}\:=\:{dx} \\ $$$$\:\:\int\:\frac{{dy}}{{y}−\mathrm{2}}−\int\frac{{dy}}{{y}+\mathrm{1}}\:=\:\int\:\mathrm{3}{dx} \\ $$$$\:\:\mathrm{ln}\:\left(\frac{{y}−\mathrm{2}}{{y}+\mathrm{1}}\right)\:=\:\mathrm{3}{x}\:+{c}\: \\ $$$$\:\:\:\frac{{y}−\mathrm{2}}{{y}+\mathrm{1}}\:=\:{Ce}^{\mathrm{3}{x}} \: \\ $$$$ \\ $$

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