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Question

Find a linearization of the differential equation

d2xdt2+xe0.01x=0.\frac{d^{2} x}{d t^{2}}+x e^{0.01 x}=0.

Solution

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The Maclaurin series for exe^x is

ex=n=0xnn!=1+x+x22+x36+e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots

For small values of xx, we can approximate xe0.01xxe^{0.01x} as

xe0.01x=x(1+0.01x+(0.01x)22+(0.01x)36+)x.xe^{0.01x} = x \left( 1 + 0.01x + \frac{(0.01x)^2}{2} + \frac{(0.01x)^3}{6} + \cdots \right) \approx x.

We used the first term in the series expansion of e0.01xe^{0.01x}. We could not use more terms, because then we wouldn't get a linear equation. The linearization of the equation is then

d2xdt2+x=0\color{#4257b2} \boxed{ \frac{d^2x}{dt^2} + x = 0 }

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