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

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