Let y=y(x) be the solution curve of the differential equation, (y2−x)dydx=1, satisfying y(0)=1. This curve intersects the x-axis at a point whose abscissa is :
A
2+e
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B
2
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C
2−e
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D
−e
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Solution
The correct option is C2−e (y2−x)dydx=1 ⇒dxdy+x=y2 I.F.=e∫1dy=ey
Therefore, the solution curve is xey=∫y2eydy ⇒x=y2−2y+2+ce−y
Given y(0)=1 ⇒c=−e ∴ Solution is x=y2−2y+2−e−y+1 ∴ The value of x where the curve cuts the x-axis is x=2−e