Let y = y(x) be the solution curve of the differential equation, (y2-x) (dy/dx)=1, satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is :

Question

Let y = y(x) be the solution curve of the differential equation, (y2-x) (dy/dx)=1, satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is : (A) 2 + e (B) 2 (C) 2 - e (D) -e

Tardigrade Admin · Accepted Answer

(y2-x) (dy/dx)=1 ⇒ (dx/dy)+x=y2 I.F.=e∫ dy=ey Solution is given by x ey=(y2-2y+2)ey+C x = 0, y = 1, gives C = -e If y = 0, then x = 2 - e

Q. Let $y = y (x)$ be the solution curve of the differential equation, $(y^{2} - x) \frac{d y}{d x} = 1,$ satisfying $y (0) = 1$ . This curve intersects the x-axis at a point whose abscissa is :

$2 + e$

$2$

$2 - e$

$- e$

$(y^{2} - x) \frac{d y}{d x} = 1$
$\Rightarrow \frac{d x}{d y} + x = y^{2}$
$I . F . = e^{\int d y} = e^{y}$
Solution is given by
$x e^{y} = (y^{2} - 2 y + 2) e^{y} + C$
$x = 0, y = 1$ , gives $C = - e$
If $y = 0$ , then $x = 2 - e$

Q. Let y=y(x) be the solution curve of the differential equation, (y2−x)dxdy​=1, satisfying y(0)=1. This curve intersects the x-axis at a point whose abscissa is :

2+e

2

2−e

−e

(y2−x)dxdy​=1 ⇒dydx​+x=y2 I.F.=e∫dy=ey Solution is given by xey=(y2−2y+2)ey+C x=0,y=1, gives C=−e If y=0, then x=2−e

Q. Let $y = y (x)$ be the solution curve of the differential equation, $(y^{2} - x) \frac{d y}{d x} = 1,$ satisfying $y (0) = 1$ . This curve intersects the x-axis at a point whose abscissa is :

$2 + e$

$2$

$2 - e$

$- e$

$(y^{2} - x) \frac{d y}{d x} = 1$
$\Rightarrow \frac{d x}{d y} + x = y^{2}$
$I . F . = e^{\int d y} = e^{y}$
Solution is given by
$x e^{y} = (y^{2} - 2 y + 2) e^{y} + C$
$x = 0, y = 1$ , gives $C = - e$
If $y = 0$ , then $x = 2 - e$