Let y - y x be the solution curve of the differential equation- y 2- x dy - dx -1- satisfying y 0-1- This curve intersects the x-axis at a point whose abscissa is -A- - eB- 2C- 2-eD- 2- e

The correct option is D 2 e(y^2 x)dydx=1 ⇒dxdy+x=y^2 I.F.=e^∫1dy=e^y Therefore, the solution curve is xe^y=∫ y^2 e^y dy ⇒ x=y^2 2y+2+ce^ y Given y(0) = 1 ⇒ ...

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Standard XII

Mathematics

Solving Linear Differential Equations of First Order

Let y = y x b...

Question

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

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2

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2 - e

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- e

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Solution

The correct option is C $2 - e$
$(y^{2} - x) \frac{d y}{d x} = 1$
$\Rightarrow \frac{d x}{d y} + x = y^{2}$
$I.F. = e^{\int 1 d y} = e^{y}$

Therefore, the solution curve is
$x e^{y} = \int y^{2} e^{y} d y$
$\Rightarrow x = y^{2} - 2 y + 2 + c e^{- y}$
Given $y (0) = 1$
$\Rightarrow c = - e$
$∴$ Solution is $x = y^{2} - 2 y + 2 - e^{- y + 1}$
$∴$ The value of $x$ where the curve cuts the $x$ -axis is $x = 2 - e$

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Solving Linear Differential Equations of First Order

Standard XII Mathematics

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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 :

The correct option is C 2−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

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 :