Question

Linearizing differential equation

Answer 1

what does linearize mean? to me this ODE is already linear as it is

Answer 2

\[\frac{ d^3\delta x+0 }{ dt^3 } +10*\frac{ d^2 \delta x +0}{ dt^2 } +20*\frac{ d \delta x +0}{ dt } +15x = f(x)\]

Answer 3

\[\frac{ d^3\delta x }{ dt^3 } +10*\frac{ d^2 \delta x }{ dt^2 } +20*\frac{ d \delta x }{ dt } +15 \delta x = f(x)\]

Answer 4

\[15 \delta x - 15(0) = \frac{ d }{ dx }(15 x) |x=0 * \delta f(x)\]

Answer 5

where ,
\[\delta f(x) = 3e ^{-5x} - 3e^0\]

Answer 6

\[15 \delta x - 0 = 15(3e^{-5x} + 3) \]

Answer 7

\[\frac{ d^3\delta x }{ dt^3 } +10*\frac{ d^2 \delta x }{ dt^2 } +20*\frac{ d \delta x }{ dt } +15 \delta x = 0\]

Answer 8

Now substitute ?

Answer 9

\[\frac{ d^3\delta x }{ dt^3 } +10*\frac{ d^2 \delta x }{ dt^2 } +20*\frac{ d \delta x }{ dt } +15(3e^{-5x}+3) = 0\]

Answer 10

\[\frac{ d^3\delta x }{ dt^3 } +10*\frac{ d^2 \delta x }{ dt^2 } +20*\frac{ d \delta x }{ dt } +15 *3e^{-5x} = -15*3\]

Answer 11

Its still not linear due to e^-5x ? I have no idea what to do or whether I am doing this correctly . Please Help

Answer 12

@pooja195

Answer 13

you have to turn the expression \(3 e^{- 5x}\) into the form \( a x + b\)

all that \(\delta\) stuff on the LHS of your equation is a side show, so i'd forget about it until you have linearised.

you can do that by trotting out the Taylor Series expansion for \(e^z\)

then your DE will be linear too.