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Question Number 96596 by bemath last updated on 03/Jun/20

$$x(x+1)\frac{dy}{dx}-(2x+1)y=0$$

Answered by bobhans last updated on 03/Jun/20

$$\frac{dy}{dx}=\frac{2x+1}{x^2+x}.y\Rightarrow \frac{dy}{y}=\frac{(2x+1)dx}{x^2+x}$$$$\ln \mid \mathrm{y}\mid =\ln \mid \mathrm{C}(x^2+x)\mid \Rightarrow \mathrm{y}=\mathrm{C}(x^2+x)$$

Answered by mathmax by abdo last updated on 03/Jun/20

$$\mathrm{x}(\mathrm{x}+1){\mathrm{y}}^{{}^{\prime }}-(2\mathrm{x}+1)\mathrm{y}=0\Rightarrow \mathrm{x}(\mathrm{x}+1){\mathrm{y}}^{{}^{\prime }}=(2\mathrm{x}+1)\mathrm{y}\Rightarrow \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=\frac{2\mathrm{x}+1}{\mathrm{x}(\mathrm{x}+1)}$$$$\Rightarrow \ln \mid \mathrm{y}\mid =\int \frac{2\mathrm{x}+1}{\mathrm{x}(\mathrm{x}+1)}\mathrm{dx}+\mathrm{c}=\int (\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}+1})\mathrm{dx}+\mathrm{c}=\ln \mid \mathrm{x}\mid +\ln \mid \mathrm{x}+1\mid +\mathrm{c}\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=\mathrm{k}\mid \mathrm{x}\mid \times \mid \mathrm{x}+1\mid =\mathrm{k}\mid {\mathrm{x}}^2+\mathrm{x}\mid$$$$$$

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