Question

Question: Identify each of the differential equations in Problems 1to 24 as to type (for example, separable, linear first order, linear second order, etc.), and then solve it.

$\mathit{u}\mathbf{(}\mathbf{1}\mathbf{-}\mathit{v}\mathbf{)}\mathit{d}\mathit{v}\mathbf{+}{\mathit{v}}^{\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{-}\mathit{u}\mathbf{)}\mathit{d}\mathit{u}\mathbf{=}\mathbf{0}$

Short answer

The solution of given differential equation is $u-\ln u+\ln v+v^{-1}=C$.

Step-by-step solution

Given information.

The differential equation is $u(1-v)dv+v^2(1-u)du=0$.

Differential equation.

When fand its derivatives are inserted into the equation, a solution is a function y = f(x) that solves the differential equation. The highest order of any derivative of the unknown function appearing in the equation is the order of a differential equation.

A differential equation of the form$\mathbf{(}\mathit{D}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{b}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{a}\mathbf{\ne }\mathit{b}$ has general solution$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{a}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{b}\mathbf{x}}\mathbf{.}$

Find the solution of the given differential equation.

Consider the equation.

$u(1-v)dv+v^2(1-u)du=0$

Rearrange above equation.

$\begin{array}{c}u(1-v)dv=-v^2(1-u)du\\ \frac{(1-v)}{v^2}dv=-\frac{(1-u)}{u}du\\ \left(\frac{1}{v^2}-\frac{1}{v}\right)dv=\frac{(u-1)}{u}du\end{array}$

The above equation it is clear that the equation can be solved by separation of variables.

Integrate above equation.

$\begin{array}{c}\int \left(\frac{1}{v^2}-\frac{1}{v}\right)dv=\int \frac{(u-1)}{u}du\\ v^{-1}-\ln v=u-\ln u-C\\ u-\ln u+\ln v+v^{-1}=C\end{array}$

Thus, the solution of given differential equation is$u-\ln u+\ln v+v^{-1}=C$ .