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Question Number 133570 by Ahmed1hamouda last updated on 23/Feb/21

Commented by Ahmed1hamouda last updated on 23/Feb/21

Solve the differential equations

Commented by EDWIN88 last updated on 23/Feb/21

$$$$$$\mathrm{fragst}\mathrm{du}\mathrm{ernsthaft}$$

Answered by EDWIN88 last updated on 23/Feb/21

$$\mathrm{Solve}\mathrm{the}\mathrm{differential}\mathrm{equation}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{3\mathrm{x}+2\mathrm{y}-5}{2\mathrm{x}+3\mathrm{y}-5}=0$$$$\mathrm{set}\{\begin{array}{l}\mathrm{x}={\mathrm{x}}_1+\mathrm{h}\\ \mathrm{y}={\mathrm{y}}_1+\mathrm{k}\end{array}$$$$\frac{{\mathrm{dy}}_1}{{\mathrm{dx}}_1}=\frac{-{3\mathrm{x}}_1-{2\mathrm{y}}_1-3\mathrm{h}-2\mathrm{k}+5}{{2\mathrm{x}}_1+{3\mathrm{y}}_1+2\mathrm{h}+3\mathrm{k}+5}$$$$\mathrm{solving}\mathrm{set}\mathrm{of}\mathrm{two}\mathrm{equation}\{\begin{array}{l}-3\mathrm{h}-2\mathrm{k}+5=0\\ 2\mathrm{h}+3\mathrm{k}+5=0\end{array}$$$$\mathrm{we}\mathrm{find}\mathrm{h}=1\mathrm{and}\mathrm{k}=1$$$$\mathrm{the}\mathrm{result}\frac{{\mathrm{dy}}_1}{{\mathrm{dx}}_1}=\frac{-{3\mathrm{x}}_1-{2\mathrm{y}}_1}{{2\mathrm{x}}_1+{3\mathrm{y}}_1};\mathrm{let}{\mathrm{y}}_1={\mathrm{ux}}_1$$$$\Rightarrow \mathrm{u}+{\mathrm{x}}_1\frac{\mathrm{du}}{{\mathrm{dx}}_1}=\frac{-{3\mathrm{x}}_1-{2\mathrm{ux}}_1}{{2\mathrm{x}}_1+{3\mathrm{ux}}_1}=\frac{-3-2\mathrm{u}}{2+3\mathrm{u}}$$$$\Rightarrow {\mathrm{x}}_1\frac{\mathrm{du}}{{\mathrm{dx}}_1}=\frac{-3-2\mathrm{u}-2\mathrm{u}-{3\mathrm{u}}^2}{2+3\mathrm{u}}$$$$\Rightarrow \frac{(2+3\mathrm{u})}{{3\mathrm{u}}^2+4\mathrm{u}+3}\mathrm{du}=-\frac{{\mathrm{dx}}_1}{{\mathrm{x}}_1}$$$$\Rightarrow \int \frac{\mathrm{d}({3\mathrm{u}}^2+4\mathrm{u}+3)}{{3\mathrm{u}}^2+4\mathrm{u}+3}=-2\int \frac{{\mathrm{dx}}_1}{{\mathrm{x}}_1}$$$$\ln \mid {3\mathrm{u}}^2+4\mathrm{u}+3\mid =\ln \mid \frac{1}{{\mathrm{x}}_1^2}\mid +\mathrm{c}$$$$\Rightarrow {3\mathrm{u}}^2+4\mathrm{u}+3=\frac{\mathrm{C}}{{\mathrm{x}}_1^2}$$$$\Rightarrow 3{\left(\frac{{\mathrm{y}}_1}{{\mathrm{x}}_1}\right)}^2+4\left(\frac{{\mathrm{y}}_1}{{\mathrm{x}}_1}\right)+3=\frac{\mathrm{C}}{{\mathrm{x}}_1^2}$$$$\Rightarrow {3\mathrm{y}}_1^2+{4\mathrm{x}}_1{\mathrm{y}}_1+{3\mathrm{x}}_1^2=\mathrm{C}$$$$\Rightarrow 3{(\mathrm{y}-1)}^2+4(\mathrm{x}-1)(\mathrm{y}-1)+3{(\mathrm{x}-1)}^2=\mathrm{C}$$

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