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 $\dfrac{1}{x}dx=dt$.