JEE MAIN - Mathematics (2019 - 12th January Morning Slot - No. 23)
Let y = y(x) be the solution of the differential equation, x$${{dy} \over {dx}}$$ + y = x loge x, (x > 1). If 2y(2) = loge 4 $$-$$ 1, then y(e) is equal to :
$$ - {e \over 2}$$
$$ - {{{e^2}} \over 2}$$
$${{{e^2}} \over 4}$$
$${e \over 4}$$
Explanation
$${{dy} \over {dx}} = {y \over x} = \ell nx$$
$${e^{\int {{1 \over x}dx} }} = x$$
$$xy = \int {x\ell nx + C} $$
$$\ell nx{{{x^2}} \over 2} - \int {{1 \over x}.{{{x^2}} \over 2}} $$
$$xy = {x \over 2}\ell nx - {{{x^2}} \over 4} + C,$$
for $$2y\left( 2 \right) = 2\ell n2 - 1$$
$$ \Rightarrow $$ $$C = 0$$
$$y = {x \over 2}\ell nx - {x \over 4}$$
$$y\left( e \right) = {e \over 4}$$
$${e^{\int {{1 \over x}dx} }} = x$$
$$xy = \int {x\ell nx + C} $$
$$\ell nx{{{x^2}} \over 2} - \int {{1 \over x}.{{{x^2}} \over 2}} $$
$$xy = {x \over 2}\ell nx - {{{x^2}} \over 4} + C,$$
for $$2y\left( 2 \right) = 2\ell n2 - 1$$
$$ \Rightarrow $$ $$C = 0$$
$$y = {x \over 2}\ell nx - {x \over 4}$$
$$y\left( e \right) = {e \over 4}$$
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