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Mathematics Question on Differential equations

Let y = y(x) be the solution curve of the differential equation

dydx+1x2−1y=(x−1x+1)1/2,x>1\begin{array}{l} \frac{dy}{dx}+\frac{1}{x^2-1}y=\left(\frac{x-1}{x+1}\right)^{1/2},x>1 \end{array}

passing through the point (2, √(1/3)). Then √7 y(8) is

A

11+6loge311+6log_e3

B

19

C

12−2loge312-2log_e3

D

19−6loge319-6log_e3

Answer

19−6loge319-6log_e3

Explanation

Solution

dydx+1x2−1y=x−1x+1,x>1\begin{array}{l} \frac{dy}{dx}+\frac{1}{x^2-1}y=\sqrt{\frac{x-1}{x+1}},x>1\end{array}

Integrating factor I.F.=e∫1x2−1dx=e12In∣x−1x+1∣\begin{array}{l} \text{I.F.}=e^{\int\frac{1}{x^2-1}dx}=e^{\frac{1}{2}\text{In}\left|\frac{x-1}{x+1}\right|} \end{array}

Solution of differential equation yx−1x+1=∫x−1x+1dx=∫(1−2x+1)dx\begin{array}{l} y\sqrt{\frac{x-1}{x+1}}=\int\frac{x-1}{x+1}dx =\int\left(1-\frac{2}{x+1}\right)dx\end{array}

yx−1x+1=x−2In∣x+1∣+C\begin{array}{l} y\sqrt{\frac{x-1}{x+1}}=x-2\text{In}\left|x+1\right|+C\end{array}

Curve passes through (2, √ (1/3) )

13×13=2−2In3+C\begin{array}{l} \frac{1}{\sqrt{3}}\times\frac{1}{\sqrt{3}}=2-2\text{In}3+C \end{array}

C=2In3−53\begin{array}{l} C=2\text{In}3-\frac{5}{3} \end{array}

y(8)×73=8−2In9+2In3−53\begin{array}{l} y\left(8\right)\times\frac{\sqrt{7}}{3}=8-2\text{In}9+2\text{In}3-\frac{5}{3}\end{array}

7⋅y(8)=19−6In3\begin{array}{l} \sqrt{7}\cdot y\left(8\right)=19-6\text{In}3 \end{array}