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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 \((6)\).

\((sin\theta )y''' - (cos\theta )y' = 2\)

Short Answer

Expert verified

The equation is linear and third 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 matches the form, \({a_3}(\theta )\frac{{{d^3}y}}{{d{\theta ^3}}} + {a_2}(\theta )\frac{{{d^2}y}}{{d{\theta ^2}}} + {a_1}(\theta )\frac{{dy}}{{d\theta }} + {a_0}(\theta )y = g(\theta )\), then it is linear with the parameters equal to,

\(\begin{array}{l}{a_3} = sin\theta \\{a_2} = 0\\{a_1} = - cos\theta \\{a_0} = 0\end{array}\)

And\(g(\theta ) = 0\).

Also, as\(n = 3\), then it is a third order differential equation.

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