Which of the following is true for y(x) that satisfies the differential equation
${{dy} \over {dx}}$ = xy $-$ 1 + x $-$ y; y(0) = 0 :
Solution
${{dy} \over {dx}} = (x - 1)y + (x - 1)$<br><br>${{dy} \over {dx}} = (x - 1)(y + 1)$<br><br>${{dy} \over {y + 1}} = (x - 1)dx$<br><br>Integrating both sides, we get<br><br>$\ln (y + 1) = {{{x^2}} \over 2} - x + c$<br><br>$x = 0,y = 0$<br><br>$\Rightarrow c = 0$<br><br>$\therefore$ $\ln (y + 1) = {{{x^2}} \over 2} - x$<br><br>putting $x = 1,\ln (y + 1) = {1 \over 2} - 1 = - {1 \over 2}$<br><br>$y + 1 = {e^{ - {1 \over 2}}}$<br><br>$y = {e^{ - {1 \over 2}}} - 1$<br><br>$\therefore$ $y(1) = {e^{ - {1 \over 2}}} - 1$
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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