Let y = y(x) be the solution of the differential equation, x[dy dx] + y = x loge x, (x 1…

Question

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 :

Accepted Answer

$${{dy} \over {dx}} = {y \over x} = \ell nx$$

$${e^{\int {{1 \over x}dx} }} = x$$

$$xy = \int {x\ell nx + C} $$

$$\ell nx{{{x^2}} \over 2} - \int {{1 \over x}.{{{x^2}} \over 2}} $$

$$xy = {x \over 2}\ell nx - {{{x^2}} \over 4} + C,$$

for   $$2y\left( 2 \right) = 2\ell n2 - 1$$

$$ \Rightarrow $$  $C = 0$

$$y = {x \over 2}\ell nx - {x \over 4}$$

$$y\left( e \right) = {e \over 4}$$

Solutions · 1