Question

The integrating factor for the differential equation \(\frac{{{\rm{dP}}}}{{{\rm{dt}}}} + {{\rm{K}}_2}{\rm{P}} = {{\rm{K}}_1}{{\rm{L}}_0}{{\rm{e}}^{ - {{\rm{K}}_1}{\rm{t}}}}\)
1. e^-k₁t
2. e^-k₂t
3. e^k₁t
4. e^k₂t

Answer 1

Correct Answer - Option 4 : e^k₂t

Concept: 

The standard form of a linear equation of the first order, commonly known as Leibnitz's linear equation, is

\({\frac{{{\rm{dy}}}}{{{\rm{dx}}}}{\rm{ + P}}{\rm{.y = Q}}}\)

Where, P and Q are the functions of 'X'

The integrating factor is given as,

\({\rm{I}}{\rm{.F}} = {{\rm{e}}^{{\rm{\smallint P}}{\rm{dx}}}}\)

The solution of the above equation is given as,

\(\begin{array}{*{20}{l}} {\left( {{\rm{I}}{\rm{.F}}} \right){\rm{y}} = {\rm{ \smallint}}\left( {{\rm{I}}{\rm{.F}}} \right){\rm{Q}}\ {\rm{dx}}}+C \end{array}\)

Calculation:

Given equation,

\(\frac{{{\rm{dP}}}}{{{\rm{dt}}}} + {{\rm{K}}_2}{\rm{P}} = {{\rm{K}}_1}{{\rm{L}}_0}{{\rm{e}}^{ - {{\rm{K}}_1}{\rm{t}}}}\)

This equation is a linear equation of first order, therefore comparing it with the general equation

\({\frac{{{\rm{dy}}}}{{{\rm{dx}}}}{\rm{ + P}}{\rm{.y = Q}}}\)

Hence P = K₂

∴ \({\rm{I}}{\rm{.F}} = {{\rm{e}}^{{\rm{\smallint P}}{\rm{dt}}}}\)

∴ \({\rm{I}}{\rm{.F}} = {{\rm{e}}^{{\rm{\smallint K_2}}{\rm{dt}}}}\)

∴ \({\rm{I}}{\rm{.F}} = {{\rm{e}}^{{\rm{ K_2}}{\rm{t}}}}\)