Question

In Problemsanddetermine whether the given first-order differential equation is linear in the indicated dependent variable by matching it with the first differential equation given in.

in; in.

Short answer

The equation is nonlinearin\(u\), and linear in\(v\).

Step-by-step solution

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,

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

Determine whether it is nonlinear

As, by the classification of linearity, the given differential equation should be in the form, \({a_1}(v)\frac{{du}}{{dv}} + {a_0}(v)u = g(v)\).

Simplify the given equation to get,

\(\begin{aligned}{c}u + (v + uv - u{e^u})\frac{{du}}{{dv}} &= 0\\(v + uv - u{e^u})\frac{{du}}{{dv}} + u &= 0\end{aligned}\)

But the term\({a_1}\) has dependence on both \(u\) and \(v\). So, the given is nonlinear in\(u\).

Determine whether it is linear.

As, by the classification of linearity, the given differential equation should be in the form,\({a_1}(u)\frac{{dv}}{{du}} + {a_0}(u)v = g(u)\).

Simplify the given equation to get,

\(\begin{aligned}{c}u\frac{{dv}}{{du}} + (v + uv - u{e^u}) &= 0\\u\frac{{dv}}{{du}} + (1 + u)v &= u{e^u}\end{aligned}\)

It is linear in \(v\) with the parameters equal to,

\(\begin{aligned}{a_1} &= u\\{a_0} &= 1 + u\end{aligned}\)

And\(g(u) = u{e^u}\).