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

Discussion Forum : Differential Equations

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

Options:

A .

y = k

B .

y = k x

C .

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

D .

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

Answer: Option D
:
D

\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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