QJEEM20Maa297: JEE Main 2020 Mathematics - Class 12 JEEM20Maa297 Solution: Let y = y(x) be the solution curve of the differential…

Question

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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} }}

= e^y

Solution is given by

xe^y =

{\int {{y^2}{e^y}dy} }

\Rightarrow

xe^y = (y² – 2y + 2)e^y + C

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

\therefore

C = -e

\therefore

xe^y = (y² – 2y + 2)e^y - e

put y = 0

\therefore

x = 0 – 0 + 2 – e

\Rightarrow

x = 2 - e

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