Let y=y(x) be the solution curve of the differential equation \frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{1 / 2}, x 1 passing through the point \left(2, \sqrt{\frac{1}{3}}\right). Then \sqrt{7}\, y(8) is equal to :

Question

Let y=y(x) be the solution curve of the differential equation \frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{1 / 2}, x >1 passing through the point \left(2, \sqrt{\frac{1}{3}}\right). Then \sqrt{7}\, y(8) is equal to :

Accepted Answer

Answer: D. 19-6 \log _{\mathrm{e}} 3 {{dy} \over {dx}} + {1 \over {{x^2} - 1}}y = \sqrt {{{x - 1} \over {x + 1}}} ,\,x > 1 Integrating factor I.F. = {e^{\int {{1 \over {{x^2} - 1}}dx} }} = {e^{{1 \over 2}\ln \left| {{{x - 1} \over {x + 1}}} ight|}} = \sqrt {{{x - 1} \over {x + 1}}} Solution of differential equation y\sqrt {{{x - 1} \over {x + 1}}} = \int {{{x - 1} \over {x + 1}}dx = \int {\left( {1 - {2 \over {x + 1}}} ight)dx} } y\sqrt {{{x - 1} \over {x + 1}}} = x - 2\ln |x + 1| + C Curve passes through \left( {2,\sqrt {{1 \over 3}} } ight) {1 \over {\sqrt 3 }} 	imes {1 \over {\sqrt 3 }} = 2 - 2\ln 3 + C C = 2\ln 3 - {5 \over 3} y(8) 	imes {{\sqrt 7 } \over 3} = 8 - 2\ln 9 + 2\ln 3 - {5 \over 3} \sqrt 7 \,.\,y(8) = 19 - 6\ln 3

Let $$y=y(x)$$ be the solution curve of the differential equation $$ \frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{1 / 2}$$, $$x >1$$ passing through the point $$\left(2, \sqrt{\frac{1}{3}}\right)$$. Then $$\sqrt{7}\, y(8)$$ is equal to :

Answer

Explanation

More JEE Math questions