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

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

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

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

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