JEE MAIN - Mathematics (2020 - 7th January Evening Slot - No. 17)
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 :
$$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$$, satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is :
2 + e
-e
2
2 - e
Explanation
$$\left( {{y^2} - x} \right){{dy} \over {dx}} = 1$$
$$ \Rightarrow $$ $${{dx} \over {dy}}$$ + 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
$$ \Rightarrow $$ $${{dx} \over {dy}}$$ + 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
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