The family of curves passing through (0,0) and satisfying the differential equation y 2 y 1 = 1 (where y n = d n y d x n ) is Options: A . &nbspy =k B . &nbspy =kx C . &nbspy=k(ex+1) D . &nbspy=k(ex−1)

Answer: Option D : D d p d x = P ( where p = d y d x ) ln P = x + c ⇒ p = e x + c d y d x = k e x y = k e x + λ Satisfying (0,0), So λ = − k y = k ( e x − 1 )

The family of curves passing through (0,0) and satisfying the differential equat

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}})

Options:

A . y =k

B . y =kx

C . y=k(ex+1)

D . y=k(ex−1)

Answer: Option D
:
D

\frac{d p}{d x} = P (where p = \frac{d y}{d x})

P = x + c \Rightarrow p = e^{x + c}

\frac{d y}{d x} = k e^{x} y = k e^{x} + λ

Satisfying (0,0), So

λ = - k

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

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