Question Number 27031 by cmaxamuud98 @gmail.com last updated on 01/Jan/18
$${xy}=\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\frac{{dy}}{{dx}}\:\:\:\:{x}=\mathrm{0}\:{y}=\mathrm{1} \\ $$
Answered by mrW1 last updated on 01/Jan/18
$$\frac{{dy}}{{y}}=\frac{{xdx}}{\mathrm{1}−{x}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{2}}×\frac{{d}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\mathrm{ln}\:{y}=−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{ln}\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)+{C} \\ $$$$\mathrm{ln}\:\mathrm{1}=−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{ln}\:\left(\mathrm{1}−\mathrm{0}\right)+{C} \\ $$$$\Rightarrow{C}=\mathrm{0} \\ $$$$\mathrm{2}\:\mathrm{ln}\:{y}=−\:\mathrm{ln}\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\left(\mathrm{1}−{x}^{\mathrm{2}} \right){y}^{\mathrm{2}} =\mathrm{1} \\ $$
Commented by abdo imad last updated on 02/Jan/18
$${and}\:{if}\:\:{x}>\mathrm{1}\:\:{what}\:{is}\:{happen}… \\ $$
Commented by prakash jain last updated on 02/Jan/18
$$\int\frac{\mathrm{1}}{\mathrm{1}−{x}}{dx}=−\mathrm{ln}\:\left(\mathrm{1}−{x}\right)+{C}_{\mathrm{1}} \\ $$$${or}−\int\frac{{dx}}{{x}−\mathrm{1}}=−\mathrm{ln}\:\left({x}−\mathrm{1}\right)+{C}_{\mathrm{2}} \\ $$$${we}\:{can}\:{always}\:{write}\:{the}\:{following}\:{since}\:{C} \\ $$$${is}\:{arbitrary}\:{constant}. \\ $$$$\Rightarrow\int\frac{\mathrm{1}}{{x}}{dx}=\mathrm{ln}\:\mid{x}\mid+{C} \\ $$