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Question

Question: The degree of the differential equation \(y ( x ) = 1 + \frac { d y } { d x } + \frac { 1 } { 1.2 ...

The degree of the differential equation

y(x)=1+dydx+11.2(dydx)2+11.2.3(dydx)3+y ( x ) = 1 + \frac { d y } { d x } + \frac { 1 } { 1.2 } \left( \frac { d y } { d x } \right) ^ { 2 } + \frac { 1 } { 1.2 .3 } \left( \frac { d y } { d x } \right) ^ { 3 } + \ldots is

A

2

B

3

C

1

D

None of these

Answer

1

Explanation

Solution

y=1+t+t22!+t33!++y = 1 + t + \frac { t ^ { 2 } } { 2 ! } + \frac { t ^ { 3 } } { 3 ! } + \ldots + \infty where t=dydxt = \frac { d y } { d x }

y=ety = e ^ { t } , ∴ t=logyt = \log ydydx=logy\frac { d y } { d x } = \log y.

Hence degree is 1.