JEE MAIN - Mathematics (2021 - 25th July Evening Shift - No. 17)
Let y = y(x) be the solution of the differential
equation xdy = (y + x3 cosx)dx with y($$\pi$$) = 0, then $$y\left( {{\pi \over 2}} \right)$$ is equal to :
equation xdy = (y + x3 cosx)dx with y($$\pi$$) = 0, then $$y\left( {{\pi \over 2}} \right)$$ is equal to :
$${{{\pi ^2}} \over 4} + {\pi \over 2}$$
$${{{\pi ^2}} \over 2} + {\pi \over 4}$$
$${{{\pi ^2}} \over 2} - {\pi \over 4}$$
$${{{\pi ^4}} \over 4} - {\pi \over 2}$$
Explanation
$$xdy = (y + {x^3}\cos x)dx$$
$$ \Rightarrow $$ $$xdy = ydx + {x^3}\cos xdx$$
$$ \Rightarrow $$ $${{xdy - ydx} \over {{x^2}}} = {{{x^3}coxdx} \over {{x^2}}}$$
$$ \Rightarrow $$ $${d \over {dx}}\left( {{y \over x}} \right) = \int {x\cos xdx} $$
$$ \Rightarrow {y \over x} = x\sin x - \int {1.\sin xdx} $$
$$ \Rightarrow $$ $${y \over x} = x\sin x + \cos x + C$$
$$ \Rightarrow 0 = - 1 + C \Rightarrow C = 1,x = \pi ,y = 0$$
so, $${y \over x} = x\sin x + \cos x + 1$$
$$y = {x^2}\sin x + x\cos x + x$$
$$x = {\pi \over 2}$$
$$y\left( {{\pi \over 2}} \right) = {{{\pi ^2}} \over 4} + {\pi \over 2}$$
$$ \Rightarrow $$ $$xdy = ydx + {x^3}\cos xdx$$
$$ \Rightarrow $$ $${{xdy - ydx} \over {{x^2}}} = {{{x^3}coxdx} \over {{x^2}}}$$
$$ \Rightarrow $$ $${d \over {dx}}\left( {{y \over x}} \right) = \int {x\cos xdx} $$
$$ \Rightarrow {y \over x} = x\sin x - \int {1.\sin xdx} $$
$$ \Rightarrow $$ $${y \over x} = x\sin x + \cos x + C$$
$$ \Rightarrow 0 = - 1 + C \Rightarrow C = 1,x = \pi ,y = 0$$
so, $${y \over x} = x\sin x + \cos x + 1$$
$$y = {x^2}\sin x + x\cos x + x$$
$$x = {\pi \over 2}$$
$$y\left( {{\pi \over 2}} \right) = {{{\pi ^2}} \over 4} + {\pi \over 2}$$
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