If a curve passes through the origin and the slope of the tangent to it at any point (x, y…

Question

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 :

Accepted Answer

Given

y (0) = 0

\& $${{dy} \over {dx}} = {{{{(x - 2)}^2} + y + 4} \over {x - 2}}$$

$$ \Rightarrow {{dy} \over {dx}} - {y \over {x - 2}} = (x - 2) + {4 \over {x - 2}}$$

$$ \Rightarrow I.F. = {e^{ - \int {{1 \over {x - 2}}dx} }} = {1 \over {x - 2}}$$

Solution of D.E.

$$ \Rightarrow y.{1 \over {x - 2}} = \int {{1 \over {x - 2}}\left( {(x - 2) + {4 \over {x - 2}}} \right)} \,.\,dx$$

$$ \Rightarrow {y \over {x - 2}} = x - {4 \over {x - 2}} + C$$

Now, at x = 0, y = 0 $$ \Rightarrow $$ C = -2

$$ \therefore $$ y = x (x - 2) - 4 - 2 (x - 2)

$$ \Rightarrow $$ y = x2 - 4x

This curve passes through (5, 5)

Solutions · 1