JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 13)
Let y = y(x) be the solution of the differential equation
$$\cos x(3\sin x + \cos x + 3)dy = (1 + y\sin x(3\sin x + \cos x + 3))dx,0 \le x \le {\pi \over 2},y(0) = 0$$. Then, $$y\left( {{\pi \over 3}} \right)$$ is equal to :
$$\cos x(3\sin x + \cos x + 3)dy = (1 + y\sin x(3\sin x + \cos x + 3))dx,0 \le x \le {\pi \over 2},y(0) = 0$$. Then, $$y\left( {{\pi \over 3}} \right)$$ is equal to :
$$2{\log _e}\left( {{{\sqrt 3 + 7} \over 2}} \right)$$
$$2{\log _e}\left( {{{3\sqrt 3 - 8} \over 4}} \right)$$
$$2{\log _e}\left( {{{2\sqrt 3 + 10} \over {11}}} \right)$$
$$2{\log _e}\left( {{{2\sqrt 3 + 9} \over 6}} \right)$$
Explanation
$$\cos x(3\sin x + \cos x + 3)dy = (1 + y\sin x(3\sin x + \cos x + 3))dx$$ ..... (1)
$$(3\sin x + \cos x + 3)(\cos x\,dy - y\sin x\,dx) = dx$$
$$\int {d(y.\cos x) = \int {{{dx} \over {3\sin x + \cos x + 3}}} } $$
$$y\cos x = \int {{1 \over {3\left( {{{2 + \tan {x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}} \right) + \left( {{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}} \right) + 3}}} $$
$$y\cos x = \int {{{{{\sec }^2}{x \over 2}} \over {6\tan {x \over 2} + 1 - {{\tan }^2}{x \over 2} + 3 + 3{{\tan }^2}{x \over 2}}}} $$
$$y\cos x = \int {{{{{\sec }^2}{x \over 2}} \over {2{{\tan }^2}{x \over 2} + 6\tan {x \over 2} + 4}}} = \int {{{{1 \over 2}{{\sec }^2}{x \over 2}dx} \over {{{\tan }^2}{x \over 2} + 3\tan {x \over 2} + 2}}} $$
$$y\cos x = \ln \left| {{{\tan {x \over 2} + 1} \over {\tan {x \over 2} + 2}}} \right| + c$$
Put x = 0 & y = 0
$$C = - \ln \left( {{1 \over 2}} \right) = \ln (2)$$
$$y\left( {{\pi \over 3}} \right) = 2\ln \left| {{{1 + \sqrt 3 } \over {1 + 2\sqrt 3 }}} \right| + \ln 2$$
$$ = 2\ln \left| {{{5 + \sqrt 3 } \over {11}}} \right| + \ln 2$$
$$ = 2\ln \left| {{{2\sqrt 3 + 10} \over {11}}} \right|$$
$$(3\sin x + \cos x + 3)(\cos x\,dy - y\sin x\,dx) = dx$$
$$\int {d(y.\cos x) = \int {{{dx} \over {3\sin x + \cos x + 3}}} } $$
$$y\cos x = \int {{1 \over {3\left( {{{2 + \tan {x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}} \right) + \left( {{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}} \right) + 3}}} $$
$$y\cos x = \int {{{{{\sec }^2}{x \over 2}} \over {6\tan {x \over 2} + 1 - {{\tan }^2}{x \over 2} + 3 + 3{{\tan }^2}{x \over 2}}}} $$
$$y\cos x = \int {{{{{\sec }^2}{x \over 2}} \over {2{{\tan }^2}{x \over 2} + 6\tan {x \over 2} + 4}}} = \int {{{{1 \over 2}{{\sec }^2}{x \over 2}dx} \over {{{\tan }^2}{x \over 2} + 3\tan {x \over 2} + 2}}} $$
$$y\cos x = \ln \left| {{{\tan {x \over 2} + 1} \over {\tan {x \over 2} + 2}}} \right| + c$$
Put x = 0 & y = 0
$$C = - \ln \left( {{1 \over 2}} \right) = \ln (2)$$
$$y\left( {{\pi \over 3}} \right) = 2\ln \left| {{{1 + \sqrt 3 } \over {1 + 2\sqrt 3 }}} \right| + \ln 2$$
$$ = 2\ln \left| {{{5 + \sqrt 3 } \over {11}}} \right| + \ln 2$$
$$ = 2\ln \left| {{{2\sqrt 3 + 10} \over {11}}} \right|$$
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