The family of curves passing through 0-0 and satisfying the differential equation y2-y1-1 where yn-dn y-d xn is A- y - kB- y - kxC- y-kex-1D- y-kex-1

The correct option is D y = k(e^x 1)dp/dx = P (where p = dy/dx) In P = x+c ⇒ p =e^x+c dy/dx = ke^X y = ke^x + λ Satisfy (0,0), So λ = k y = k(e^x 1) ...

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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 $\frac{y_{2}}{y_{1}} = 1$ (where $y_{n} = \frac{d^{n} y}{d x^{n}}$ 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 $y = k (e^{x} - 1)$
$\frac{d p}{d x} = P (w h e r e p = \frac{d y}{d x}) I n P = x + c \Rightarrow p = e^{x + c} \frac{d y}{d x} = k e^{X} y = k e^{x} + λ$
Satisfy (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 y2y1=1 (where yn=dnydxn is

The correct option is D y=k(ex−1)dpdx=P(where p=dydx)In P=x+c⇒p=ex+cdydx=keXy=kex+λ Satisfy (0,0), So λ=−ky=k(ex−1)

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

The correct option is D $y = k (e^{x} - 1)$
$\frac{d p}{d x} = P (w h e r e p = \frac{d y}{d x}) I n P = x + c \Rightarrow p = e^{x + c} \frac{d y}{d x} = k e^{X} y = k e^{x} + λ$
Satisfy (0,0), So $λ = - k y = k (e^{x} - 1)$