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In Problems \(1 - 8\) state the order of the given ordinary differential equation. Determine whether the equation is linear or nonlinear by matching it with \((6)\).

\(\ddot x - \left( {1 - \frac{{{{\dot x}^2}}}{3}} \right)\dot x + x = 0\)

Short Answer

Expert verified

The equation is nonlinear and second order.

Step by step solution

01

Classification of linearity.

If\(F\)is linear in \(y,y',...,{y^n}\), then the \({n^{th}}\)order ordinary differential equation is said to be linear. The form of the equation is given by,

\({a_n}(x)\frac{{{d^n}y}}{{d{x^n}}} + {a_{n - 1}}(x)\frac{{{d^{n - 1}}y}}{{d{x^{n - 1}}}} + L + {a_1}(x)\frac{{dy}}{{dx}} + {a_0}(x)y = g(x)\)

02

Determine whether it is linear or nonlinear.

As, by the classification of linearity, the given differential equation should be in the form,.

Simplify the given equation to get,

\(\ddot x - \dot x + \frac{{{{\dot x}^3}}}{3} + x = 0\)

But the term \({\dot x^3}\) has to be the power of \(3\). So, the given is nonlinear.

If the given equation is linear, then the term \({\dot x^3}\) must have to be the power of \(1\). Because of \(\ddot x\), the equation is second order differential equation.

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