Question

Find a linearization of the differential equation

$\frac{d^{2} x}{d t^{2}}+x e^{0.01 x}=0.$

Solution

VerifiedStep 1

1 of 2The Maclaurin series for $e^x$ is

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$

For small values of $x$, we can approximate $xe^{0.01x}$ as

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

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

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

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