The curve fx-y-0 passing through 0-2 satisfy the differential equation dydx-y3ex-y2- If the line x-ln5 intersects it at points y- and y- then the value of 2- is

Dydx=y^3e^x+y^2 ⇒ y^3dxdy=e^x+y^2 ⇒dxdye^ x 1ye^ x=1y^3 ⇒dxdye^ xy e^ x=1y^2 ⇒ddy(e^ xy)= 1y^2 On integrating, nbsp; e^ xy=1y+c The curve passes through (0,2) ...

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Byju's Answer

Standard XII

Mathematics

Stationary Point

The curve fx,...

Question

The curve $f (x, y) = 0$ passing through $(0, 2)$ satisfy the differential equation $\frac{d y}{d x} = \frac{y^{3}}{e^{x} + y^{2}} .$ If the line $x = ln 5$ intersects it at points $y = α$ and $y = β,$ then the value of $2 | α + β |$ is

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Solution

$\frac{d y}{d x} = \frac{y^{3}}{e^{x} + y^{2}}$
$\Rightarrow y^{3} \frac{d x}{d y} = e^{x} + y^{2}$
$\Rightarrow \frac{d x}{d y} e^{- x} - \frac{1}{y} e^{- x} = \frac{1}{y^{3}}$
$\Rightarrow \frac{d x}{d y} e^{- x} y - e^{- x} = \frac{1}{y^{2}}$
$\Rightarrow \frac{d}{d y} (e^{- x} y) = - \frac{1}{y^{2}}$
On integrating,
$e^{- x} y = \frac{1}{y} + c$
The curve passes through $(0, 2)$
So, $c = \frac{3}{2}$
Line $x = ln 5$ intersects curve $e^{- x} y = \frac{1}{y} + \frac{3}{2}$
$\frac{y}{5} = \frac{1}{y} + \frac{3}{2}$
$\Rightarrow 2 y^{2} - 15 y - 10 = 0$
$∴ 2 (α + β) = 15$

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The curve f(x,y)=0 passing through (0,2) satisfy the differential equation dydx=y3ex+y2. If the line x=ln5 intersects it at points y=α and y=β, then the value of 2|α+β| is

dydx=y3ex+y2 ⇒y3dxdy=ex+y2 ⇒dxdye−x−1ye−x=1y3 ⇒dxdye−xy−e−x=1y2 ⇒ddy(e−xy)=−1y2 On integrating, e−xy=1y+c The curve passes through (0,2) So, c=32 Line x=ln5 intersects curve e−xy=1y+32 y5=1y+32 ⇒2y2−15y−10=0 ∴2(α+β)=15

The curve $f (x, y) = 0$ passing through $(0, 2)$ satisfy the differential equation $\frac{d y}{d x} = \frac{y^{3}}{e^{x} + y^{2}} .$ If the line $x = ln 5$ intersects it at points $y = α$ and $y = β,$ then the value of $2 | α + β |$ is