91Ó°ÊÓ

An explicit solution is one in which the dependent variable is expressed directly in terms of the independent variable and constants.

Let the expression be\(ln\left( {\frac{{2X - 1}}{{X - 1}}} \right) = t\).

Take exponential on both sides of the equation.

\({e^{ln\left( {\frac{{2X - 1}}{{X - 1}}} \right)}} = {e^t}\)

\(\begin{aligned}{l}\frac{{2X - 1}}{{X - 1}} &= {e^t}\\2X - 1 &= (X - 1){e^t}\end{aligned}\)

Simplify the equation by using the algebra.

\(\begin{aligned}{c}2X - X{e^t} &= 1 - {e^t}\\X &= \frac{{1 - {e^t}}}{{2 - {e^t}}}\end{aligned}\)

Let the first derivative of the above function is

\(\begin{aligned}{c}X' &= \frac{{dX}}{{dt}} &= \frac{{\left( {2 - {e^t}} \right)\left( { - {e^t}} \right) - \left( {1 - {e^t}} \right)\left( { - {e^t}} \right)}}{{{{\left( {2 - {e^t}} \right)}^2}}}\\X' &= \frac{{dX}}{{dt}} &= \frac{{ - 2{e^t} + {e^{2t}} + {e^t} - {e^{2t}}}}{{{{\left( {2 - {e^t}} \right)}^2}}}\\X' &= \frac{{dX}}{{dt}} &= \frac{{ - {e^t}}}{{{{\left( {2 - {e^t}} \right)}^2}}}\end{aligned}\)

Substitute \(y\) and \(y'\) into the left-hand side of the differential equation.

\(\begin{aligned}{c}\frac{{ - {e^t}}}{{{{\left( {2 - {e^t}} \right)}^2}}} &= \left( {\frac{{ - 1}}{{2 - {e^t}}}} \right)\left( {\frac{{{e^t}}}{{2 - {e^t}}}} \right)\\\frac{{ - {e^t}}}{{{{\left( {2 - {e^t}} \right)}^2}}} &= \left( {\frac{{1 - {e^t}}}{{2 - {e^t}}} - 1} \right)\left( {1 - 2\left( {\frac{{1 - {e^t}}}{{2 - {e^t}}}} \right)} \right)\\\frac{{ - {e^t}}}{{{{\left( {2 - {e^t}} \right)}^2}}} &= \frac{{ - {e^t}}}{{{{\left( {2 - {e^t}} \right)}^2}}}\end{aligned}\)

That is same as the right-hand side of the differential equation. The indicated function is an explicit solution of the given differential equation.

Hence the interval of the solution while considering the solution as a function is,

\(\begin{array}{c}2 - {e^t} \ne 0\\2 \ne {e^t}\\t \ne ln2\end{array}\)

\(I\)is \(( - \infty ,\ln 2)\) and \((\ln 2,\infty )\).

Let the graph of the expression be,