If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is ${{{x^2} - 4x + y + 8} \over {x - 2}}$, then this curve also passes through the point :
Solution
Given<br><br>y (0) = 0<br><br>& ${{dy} \over {dx}} = {{{{(x - 2)}^2} + y + 4} \over {x - 2}}$<br><br>$\Rightarrow {{dy} \over {dx}} - {y \over {x - 2}} = (x - 2) + {4 \over {x - 2}}$<br><br>$\Rightarrow I.F. = {e^{ - \int {{1 \over {x - 2}}dx} }} = {1 \over {x - 2}}$<br><br>Solution of D.E.<br><br>$$ \Rightarrow y.{1 \over {x - 2}} = \int {{1 \over {x - 2}}\left( {(x - 2) + {4 \over {x - 2}}} \right)} \,.\,dx$$<br><br>$\Rightarrow {y \over {x - 2}} = x - {4 \over {x - 2}} + C$<br><br>Now, at x = 0, y = 0 $\Rightarrow$ C = $-$2<br><br>$\therefore$ y = x (x $-$ 2) $-$ 4 $-$ 2 (x $-$ 2)<br><br>$\Rightarrow$ y = x<sup>2</sup> $-$ 4x<br><br>This curve passes through (5, 5)
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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