Let $y=y(x)$ be the solution of the differential equation $x{\log _e}x{{dy} \over {dx}} + y = {x^2}{\log _e}x,(x > 1)$. If $y(2) = 2$, then $y(e)$ is equal to
Solution
<p>$x\ln x{{dy} \over {dx}} + y = {x^2}\ln x$</p>
<p>${{dy} \over {dx}} + {1 \over {x\ln x}}\,.\,y = x$</p>
<p>If $= {e^{\int {{1 \over {x\ln x}}dx} }} = {e^{\int {{1 \over t}dt} }}$, where $t = \ln x$</p>
<p>$= {e^{\ln t}} = t = \ln x$</p>
<p>$$y.\ln x = \int {x\ln x = {{{x^2}} \over 2}\ln x - \int {{{{x^2}} \over 2}\,.\,{1 \over x}} } $$</p>
<p>$y\ln x = {{{x^2}} \over 2}\ln x - {{{x^2}} \over 4} + C$ ..... (i)</p>
<p>$y(2) = 2 \Rightarrow C = 1$</p>
<p>Putting $x = e$ in (i),</p>
<p>$y = {{{e^2}} \over 4} + 1 = {{4 + {e^2}} \over 4}$</p>
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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