JEE MAIN - Mathematics (2022 - 28th July Evening Shift - No. 11)
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 :
$$11+6 \log _{e} 3$$
19
$$12-2 \log _{\mathrm{e}} 3$$
$$19-6 \log _{\mathrm{e}} 3$$
Explanation
$${{dy} \over {dx}} + {1 \over {{x^2} - 1}}y = \sqrt {{{x - 1} \over {x + 1}}} ,\,x > 1$$
Integrating factor I.F. $$ = {e^{\int {{1 \over {{x^2} - 1}}dx} }} = {e^{{1 \over 2}\ln \left| {{{x - 1} \over {x + 1}}} \right|}}$$
$$ = \sqrt {{{x - 1} \over {x + 1}}} $$
Solution of differential equation
$$y\sqrt {{{x - 1} \over {x + 1}}} = \int {{{x - 1} \over {x + 1}}dx = \int {\left( {1 - {2 \over {x + 1}}} \right)dx} } $$
$$y\sqrt {{{x - 1} \over {x + 1}}} = x - 2\ln |x + 1| + C$$
Curve passes through $$\left( {2,\sqrt {{1 \over 3}} } \right)$$
$${1 \over {\sqrt 3 }} \times {1 \over {\sqrt 3 }} = 2 - 2\ln 3 + C$$
$$C = 2\ln 3 - {5 \over 3}$$
$$y(8) \times {{\sqrt 7 } \over 3} = 8 - 2\ln 9 + 2\ln 3 - {5 \over 3}$$
$$\sqrt 7 \,.\,y(8) = 19 - 6\ln 3$$
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