Question Number 207420 by Wuji last updated on 14/May/24
$$\mathrm{solve}\:\mathrm{the}\:\mathrm{Differential}\:\mathrm{equation} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\frac{\mathrm{x}+\mathrm{3y}}{\mathrm{2x}}\right) \\ $$
Answered by mr W last updated on 14/May/24
$${let}\:{y}={tx} \\ $$$$\frac{{dy}}{{dx}}={t}+{x}\frac{{dt}}{{dx}} \\ $$$${t}+{x}\frac{{dt}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}{t}}{\mathrm{2}} \\ $$$$\mathrm{2}{x}\frac{{dt}}{{dx}}=\mathrm{1}+{t} \\ $$$$\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}}=\frac{{dx}}{{x}} \\ $$$$\mathrm{2ln}\:\left(\mathrm{1}+{t}\right)=\mathrm{ln}\:{x}+\mathrm{ln}\:{C}_{\mathrm{1}} \\ $$$$\left(\mathrm{1}+{t}\right)^{\mathrm{2}} ={C}_{\mathrm{1}} {x} \\ $$$${t}={C}\sqrt{{x}}−\mathrm{1} \\ $$$$\Rightarrow{y}={x}\left({C}\sqrt{{x}}−\mathrm{1}\right) \\ $$