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Question Number 175464 by nadovic last updated on 31/Aug/22

$$\mathrm{Find}\mathrm{general}\mathrm{solutions}\mathrm{the}\mathrm{following}$$$$\mathrm{differential}\mathrm{equations}$$$$\left(a\right)\frac{d^{2}y}{{dx}^2}+\frac{dy}{dx}=6y$$$$\left(\mathrm{b}\right)(3x^2+y^2)dx+(x^2+y^2)dy=0$$$$$$

Answered by floor(10²Eta[1]) last updated on 31/Aug/22

$${\mathrm{y}}^{″}+{\mathrm{y}}^{\prime }-6\mathrm{y}=0$$$$\mathrm{characteristic}\mathrm{equation}:$$$${\mathrm{D}}^2+\mathrm{D}-6=0\Rightarrow \mathrm{D}=-3,2$$$$\Rightarrow \mathrm{y}={\mathrm{Ae}}^{-3\mathrm{x}}+{\mathrm{Be}}^{2\mathrm{x}}$$

Answered by floor(10²Eta[1]) last updated on 31/Aug/22

$$\mathrm{y}\left(\mathrm{x}\right)=\mathrm{xv}\left(\mathrm{x}\right)$$$${\mathrm{y}}^{\prime }\left(\mathrm{x}\right)=\mathrm{v}\left(\mathrm{x}\right)+{\mathrm{xv}}^{\prime }\left(\mathrm{x}\right)\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}$$$$\mathrm{dy}=\mathrm{vdx}+\mathrm{xdv}$$$$({3\mathrm{x}}^2+{\mathrm{x}}^2{\mathrm{v}}^2)\mathrm{dx}+({\mathrm{x}}^2+{\mathrm{x}}^2{\mathrm{v}}^2)(\mathrm{vdx}+\mathrm{xdv})$$$$({3\mathrm{x}}^2+{\mathrm{x}}^2{\mathrm{v}}^2+{\mathrm{x}}^2\mathrm{v}+{\mathrm{x}}^2{\mathrm{v}}^3)\mathrm{dx}+({\mathrm{x}}^3+{\mathrm{x}}^3{\mathrm{v}}^2)\mathrm{dv}=0$$$$(3+{\mathrm{v}}^2+\mathrm{v}+{\mathrm{v}}^3)\mathrm{dx}+(\mathrm{x}+{\mathrm{xv}}^2)\mathrm{dv}=0$$$$\frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{{\mathrm{v}}^3+{\mathrm{v}}^2+\mathrm{v}+3}{\mathrm{x}({\mathrm{v}}^2+1)}$$$$-\frac{{\mathrm{v}}^2+1}{{\mathrm{v}}^3+{\mathrm{v}}^2+\mathrm{v}+3}\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}}$$$$-\int \frac{{\mathrm{v}}^2+1}{{\mathrm{v}}^3+{\mathrm{v}}^2+\mathrm{v}+3}\mathrm{dv}=\int \frac{\mathrm{dx}}{\mathrm{x}}$$

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