Question Number 75175 by Rio Michael last updated on 08/Dec/19
$${solve}\:{the}\:{differential}\:{equation} \\ $$$$\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)\frac{{dy}}{{dx}}\:+\:\mathrm{2}{y}\:=\:\mathrm{0}\:{when}\:{y}=\mathrm{3}\:{and}\:{x}=\:\mathrm{2},{expressing} \\ $$$${your}\:{answer}\:{in}\:{the}\:{form}\:{y}={f}\left({x}\right) \\ $$
Answered by Kunal12588 last updated on 08/Dec/19
$$\left({x}^{\mathrm{2}} −\mathrm{1}\right)\frac{{dy}}{{dx}}+\mathrm{2}{y}=\mathrm{0} \\ $$$$\Rightarrow\frac{{dy}}{\mathrm{2}{y}}=−\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dy}}{{y}}=\int\frac{{dx}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }=\frac{{A}}{\left(\mathrm{1}−{x}\right)}+\frac{{B}}{\left(\mathrm{1}+{x}\right)} \\ $$$$\Rightarrow\mathrm{1}={A}\left(\mathrm{1}+{x}\right)+{B}\left(\mathrm{1}−{x}\right) \\ $$$$\Rightarrow{A}=\frac{\mathrm{1}}{\mathrm{2}},{B}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{log}\left({y}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\mathrm{1}−{x}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\mathrm{1}+{x}} \\ $$$$\Rightarrow{log}\:{y}\:=\:−{log}\left(\mathrm{1}−{x}\right)+{log}\left(\mathrm{1}+{x}\right)+{log}\:{C} \\ $$$$\Rightarrow{y}=\frac{\left(\mathrm{1}+{x}\right){C}}{\mathrm{1}−{x}} \\ $$$$\mathrm{3}=\frac{\mathrm{3}{C}}{−\mathrm{1}} \\ $$$$\Rightarrow{C}=−\mathrm{1} \\ $$$${y}=\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}} \\ $$
Commented by Rio Michael last updated on 08/Dec/19
$${thank}\:{you}\:{sir} \\ $$