The family of curves passing through 0-0 and satisfying the differential equation y - y -1- where y - dy - dx and -y - d 2 y - dx 2 i-A- y-kex-1B- y - k e x -1C- y - kD- y - kx

The correct option is B y =k(e^x 1)dpdx=p(where p =dy/dx ) ln p =x +c ⇒ p =e^x+c dydx=ke^x ⇒ y =ke^x +λ Satisfying (0,0), So λ = k y =k(e^x 1) nbsp; ...

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Standard XII

Mathematics

Variable Separable Method

The family of...

Question

The family of curves passing through $(0, 0)$ and satisfying the differential equation $\frac{y^{''}}{y^{'}} = 1$ $(where y^{'} = \frac{d y}{d x} and y^{''} = \frac{d^{2} y}{d x^{2}})$ is

y = k

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y = k x

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y = k (e^{x} + 1)

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y = k (e^{x} - 1)

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Solution

The correct option is D $y = k (e^{x} - 1)$
$\frac{d p}{d x} = p (where p = \frac{d y}{d x})$
$ln p = x + c \Rightarrow p = e^{x + c}$
$\frac{d y}{d x} = k e^{x} \Rightarrow y = k e^{x} + λ$
Satisfying $(0, 0)$ , So $λ = - k$
$y = k (e^{x} - 1)$

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The family of curves passing through (0,0) and satisfying the differential equation y′′y′=1 (where y′=dydx and y′′=d2ydx2) is

The correct option is D y=k(ex−1)dpdx=p(where p=dydx) lnp=x+c⇒p=ex+c dydx=kex⇒y=kex+λ Satisfying (0,0), So λ=−k y=k(ex−1)

The family of curves passing through $(0, 0)$ and satisfying the differential equation $\frac{y^{''}}{y^{'}} = 1$ $(where y^{'} = \frac{d y}{d x} and y^{''} = \frac{d^{2} y}{d x^{2}})$ is

The correct option is D $y = k (e^{x} - 1)$
$\frac{d p}{d x} = p (where p = \frac{d y}{d x})$
$ln p = x + c \Rightarrow p = e^{x + c}$
$\frac{d y}{d x} = k e^{x} \Rightarrow y = k e^{x} + λ$
Satisfying $(0, 0)$ , So $λ = - k$
$y = k (e^{x} - 1)$