"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 :"

Question

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"

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 :

Solution

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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 :

Solution

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More Differential Equations questions

Drill 25 more like these. Every day. Free.

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 :