Let y = y(x) be the solution curve of the differential equation, [( y^2 - x )dy dx = 1], s…

Question

Let y = y(x) be the solution curve of the differential equation,

$$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$$, satisfying y(0) =
1. This curve intersects the x-axis at a point whose abscissa is :

Accepted Answer

$$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$$

$$ \Rightarrow $$ $${{dx} \over {dy}}$$ + x = y2

I.F = $${e^{\int {dy} }}$$ = ey

Solution is given by

xey = $${\int {{y^2}{e^y}dy} }$$

$$ \Rightarrow $$ xey = (y2 -- 2y + 2)ey + C

y(0) = 1 means x = 0, y = 1

$$ \therefore $$ C = -e

$$ \therefore $$ xey = (y2 -- 2y + 2)ey - e

put y = 0

$$ \therefore $$ x = 0 -- 0 + 2 -- e

$$ \Rightarrow $$ x = 2 - e

Solutions · 1