Question

Solve the equation: $(y log x - 1)ydx = xdy$

Answer 1

- Hint: This is a linear differential equation. We can solve this by bringing it in the form-
$\dfrac{{dy}}{{dx}} + yP\left( {\text{x}} \right) = {\text{Q}}\left( {\text{x}} \right)$
Here P(x) and Q(x) are functions in x. After bringing the equation in this form, we find and multiply it with the integrating factor which is given by-
${\text{I}}.{\text{F}} = {{\text{e}}^{\smallint {\text{P}}\left( {\text{x}} \right)dx}}$
The equation can then be written in the form-
${\text{y}}\left( {{\text{I}}.{\text{F}}} \right) = \smallint {\text{Q}}\left( {\text{x}} \right).\left( {{\text{I}}.{\text{F}}} \right) + {\text{C}}$

Complete step-by-step solution -
We have been given that-
$(ylogx - 1)ydx = xdy$
Dividing the equation by dx, we get-
$\begin{aligned} & {\text{x}}\dfrac{{dy}}{{dx}} = \left( {{{\text{y}}^2}logx - {\text{y}}} \right) \\\ & \dfrac{{dy}}{{dx}} = \dfrac{{{{\text{y}}^2}}}{{\text{x}}}logx - \dfrac{{\text{y}}}{{\text{x}}} \\\ & \dfrac{{dy}}{{dx}} + \dfrac{{\text{y}}}{{\text{x}}} = {{\text{y}}^2}\dfrac{{logx}}{{\text{x}}} \\\ \end{aligned}$

We know that the term Q(x) should be independent of y, so we need to divide the equation with $y^2$ as-
$\dfrac{1}{{{{\text{y}}^2}}}\dfrac{{dy}}{{dx}} + \dfrac{1}{{xy}} = \dfrac{{logx}}{{\text{x}}}$
To simplify further, we need to make a substitution as-
${\text{t}} = \dfrac{1}{{\text{y}}}$
$dt = - \dfrac{1}{{{{\text{y}}^2}}}dy$
Substituting these values in the equation we get-
$- \dfrac{{dt}}{{dx}} + \dfrac{{\text{t}}}{{\text{x}}} = \dfrac{{logx}}{{\text{x}}}$
$\dfrac{{dt}}{{dx}} - \dfrac{{\text{t}}}{{\text{x}}} = - \dfrac{{logx}}{{\text{x}}}$

By comparison, we can clearly see that-
${\text{P}}\left( {\text{x}} \right) = - \dfrac{1}{{\text{x}}}\;and\;{\text{Q}}\left( {\text{x}} \right) = - \dfrac{{logx}}{{\text{x}}}$
We can find the integrating factor as-
$\begin{aligned} &{\text{I}}.{\text{F}} = {{\text{e}}^{\smallint {\text{P}}\left( {\text{x}} \right)dx}} = {{\text{e}}^{\smallint - \dfrac{1}{{\text{x}}}dx}} = {{\text{e}}^{ - logx}} = {{\text{e}}^{\log \left( {\dfrac{1}{{\text{x}}}} \right)}} \\\ & We\;know\;that\;{{\text{e}}^{logx}} = {\text{x}}\;so,\; \\\ & {\text{I}}.{\text{F}} = {{\text{e}}^{\log \left( {\dfrac{1}{{\text{x}}}} \right)}} = \dfrac{1}{{\text{x}}} \\\ \end{aligned}$

The equation can now be written as-
${\text{y}}\left( {{\text{I}}.{\text{F}}} \right) = \smallint {\text{Q}}\left( {\text{x}} \right).\left( {{\text{I}}.{\text{F}}} \right) + {\text{C}}$
${\text{t}}\left( {\dfrac{1}{{\text{x}}}} \right) = \smallint \left( { - \dfrac{{logx}}{{\text{x}}}} \right)\left( {\dfrac{1}{{\text{x}}}} \right)dx$
$\dfrac{{\text{t}}}{{\text{x}}} = - \smallint \dfrac{{logx}}{{{{\text{x}}^2}}}dx$
We will need to use integration by parts to further solve this problem-
$\smallint {\text{u}}\left( {\text{x}} \right){\text{v}}\left( {\text{x}} \right)dx = \left[ {{\text{v}}\left( {\text{x}} \right).\smallint {\text{u}}\left( {\text{x}} \right)dx} \right] - \smallint \left( {{\text{v}}'\left( {\text{x}} \right).\smallint {\text{u}}\left( {\text{x}} \right)dx} \right)dx$
We will assume that v(x) = logx and u(x) = x-2
$\begin{aligned} &= \smallint \dfrac{{logx}}{{{{\text{x}}^2}}}dx \\\ &= logx\smallint \dfrac{1}{{{{\text{x}}^2}}}dx - \smallint \left( {\dfrac{{d\left( {logx} \right)}}{{d{\text{x}}}}.\smallint \dfrac{1}{{{{\text{x}}^2}}}dx} \right)dx \\\ &= - \dfrac{{logx}}{{\text{x}}} - \left( {\smallint \dfrac{1}{{\text{x}}}.\left( { - \dfrac{1}{{\text{x}}}} \right)dx} \right) \\\ &= - \dfrac{{logx}}{{\text{x}}} + \smallint \dfrac{1}{{{{\text{x}}^2}}}dx = - \dfrac{{logx}}{{\text{x}}} - \dfrac{1}{{\text{x}}} + {\text{C}} \\\ \end{aligned}$

So, we can proceed as-
\dfrac{{\text{t}}}{{\text{x}}} = \dfrac{{logx}}{{\text{x}}} + \dfrac{1}{{\text{x}}} + {\text{C}} \\\
We know that ${\text{t}} = \dfrac{1}{{\text{y}}}$ so,
\dfrac{1}{{xy}} = \dfrac{{logx}}{{\text{x}}} + \dfrac{1}{{\text{x}}} + {\text{C}} \\\
Multiplying by ${\text{x}}$ and taking reciprocal we get,
${\text{y}} = \dfrac{1}{{1 + logx + Cx}} \\\$

Note: We need to know multiple concepts in this problem ranging from differential equations and calculus to the concept of logarithmic functions. We need to remember the formula for integrations and even the method to find the solution of first order linear differential equations. We should also never forget to add the constant of integration.