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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 \((6)\).

\({t^5}{y^{(4)}} - {t^3}y'' + 6y = 0\)

Short Answer

Expert verified

The equation is linear and fourth 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,\({a_4}(x)y'''' + {a_3}(x)y''' + {a_2}(x)y'' + {a_1}(x)y' + {a_0}(x)y = g(x)\), then it is linear with the parameters equal to,

\(\begin{aligned}{l}{a_4} = {t^5}\\{a_3} = 0\\{a_2} = - {t^3}\\{a_1} = 0\\{a_0} = 6\end{aligned}\)

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

Also, as \(n = 4\), then it is a fourth order differential equation.

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