Question

The integrating factor of the differential equation \[3x{{\log }_{e}}x\dfrac{dy}{dx}+y=2{{\log }_{e}}x\] is given by:
(a) \[{{\left( {{\log }_{e}}x \right)}^{2}}\]
(b) \[{{\log }_{e}}\left( {{\log }_{e}}x \right)\]
(c) \[{{\log }_{e}}x\]
(d) \[{{\left( {{\log }_{e}}x \right)}^{\dfrac{1}{3}}}\]

Answer 1

Hint: Here, we can see that the given equation is a differential equation of the form \[\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)\]. Hence our integrating factor would be \[{{e}^{\int{P\left( x \right)dx}}}\]. By substituting the value of P(x) from the given equation, find the integrating factor.

Complete step-by-step answer:

Here, we have to find the integrating factor of the differential equation \[3x{{\log }_{e}}x\dfrac{dy}{dx}+y=2{{\log }_{e}}x\].
Let us consider the differential equation given in the question: \[3x{{\log }_{e}}x\dfrac{dy}{dx}+y=2{{\log }_{e}}x\]. Let us divide the whole equation by \[3x{{\log }_{e}}x\], we get,
\[\dfrac{dy}{dx}+\dfrac{1}{3x{{\log }_{e}}x}y=\dfrac{2}{3x}....\left( i \right)\]
We know that for the general first-order differential equation, \[\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)\]. We have an integration factor \[={{e}^{\int{P\left( x \right)dx}}}\]. So by comparing equation (i) by general first-order differential equation, we get, \[P\left( x \right)=\dfrac{1}{3x{{\log }_{e}}x}\]. Hence, we get, Integration factor \[={{e}^{\int{\dfrac{1}{3x{{\log }_{e}}x}}dx}}\].
Let us assume \[\int{\dfrac{1}{3x{{\log }_{e}}x}dx=I}\]. Therefore, we get, Integration factor \[={{e}^{I}}....\left( ii \right)\].
Now, let us consider, I \[=\int{\dfrac{1}{3x{{\log }_{e}}x}dx....\left( iii \right)}\]
Let us take \[{{\log }_{e}}x=t\] and we know that \[\dfrac{d}{dx}\left( {{\log }_{e}}x \right)=\dfrac{1}{x}\]. Therefore, by differentiating both sides, we get,
\[\dfrac{1}{x}dx=dt\]
By substituting \[{{\log }_{e}}x=t\] and \[\dfrac{1}{x}dx=dt\] in equation (iii), we get,
\[I=\dfrac{1}{3}\int{\dfrac{1}{t}dt}\]
We know that \[\int{\dfrac{1}{x}dx={{\log }_{e}}x}\]. By using this, we get,
\[I=\dfrac{1}{3}\left( {{\log }_{e}}t \right)\]
We know that \[t={{\log }_{e}}x\]. Therefore, we get,
\[I=\dfrac{1}{3}\left[ {{\log }_{e}}\left( {{\log }_{e}}x \right) \right]\]
By substituting the value of I in equation (ii), we get,
Integrating factor \[={{e}^{\dfrac{1}{3}\left[ {{\log }_{e}}\left( {{\log }_{e}}x \right) \right]}}\]
We know that $a\log b=\log {{b}^{a}}$, so we can rewrite \[{{e}^{\dfrac{1}{3}\left[ {{\log }_{e}}\left( {{\log }_{e}}x \right) \right]}}\] as ${{e}^{\left[ {{\log }_{e}}{{\left( {{\log }_{e}}x \right)}^{\dfrac{1}{3}}} \right]}}$ . We also know that ${{e}^{\log x}}=x$, so we can again rewrite ${{e}^{\left[ {{\log }_{e}}{{\left( {{\log }_{e}}x \right)}^{\dfrac{1}{3}}} \right]}}$ as ${{\left( {{\log }_{e}}x \right)}^{\dfrac{1}{3}}}$ . Therefore, we get \[I.F={{\left( {{\log }_{e}}x \right)}^{\dfrac{1}{3}}}\].
Therefore our integrating factor is \[{{\left( {{\log }_{e}}x \right)}^{\dfrac{1}{3}}}\]
Hence, option (d) is the right answer.

Note: Here, students often make the mistake of writing \[\int{P\left( x \right)dx}\] as an integrating factor after finding it. But they must not forget that the integrating factor is \[{{e}^{\int{P\left( x \right)dx}}}\]. So after getting \[\int{P\left( x \right)dx}\], always remember to substitute it in the power of e to get the desired result. Also, whenever \[\dfrac{1}{x}dx\] and \[{{\log }_{e}}x\] come together, students should always remember the substitution of \[{{\log }_{e}}x=t\] and \[\dfrac{1}{x}dx=dt\].