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What is the general solution of the differential equation: $\left( {y\ln y} \right)dx - xdy = 0?$

Last updated date: 20th Jun 2024
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Hint:The given differential equation can be solved by separable method, in which we separate variables and integrate them. Also to integrate one of the integrals use substitution method and then to get an explicit general solution, take both side anti logarithms (or exponential to the power e).

Complete step by step solution:
To find the general solution of the given differential equation: $\left(
{y\ln y} \right)dx - xdy = 0$, we will use a separation method, as we can see this is a type of separable differential equation.
As the differential equation $\left( {y\ln y} \right)dx - xdy = 0$ is a separable differential equation, separating the variables in it as follows
\Rightarrow \left( {y\ln y} \right)dx - xdy = 0 \\
\Rightarrow \left( {y\ln y} \right)dx = xdy \\
\Rightarrow \dfrac{{dx}}{x} = \dfrac{{dy}}{{\left( {y\ln y} \right)}} \\
We have separated the variables, now integrating both sides, we will get
\[ \Rightarrow \int {\dfrac{1}{x}dx} = \int {\dfrac{1}{{\left( {y\ln y} \right)}}} dy\]
As we can see the left hand integral is trivial but the right hand side integral is unsetting, so we will try a substitution method to simplify it.
Let us take $u = \ln y \Rightarrow du = \dfrac{1}{y}dy$
Substituting this in above integral, we will get
\[ \Rightarrow \int {\dfrac{1}{x}dx} = \int {\dfrac{1}{u}du} \]
Now we can integrate it easily,
\[ \Rightarrow \ln x + c = \ln u\]
Using property of log, we can also write it as
\Rightarrow \ln x + \ln {\text{C}} = \ln u \\
\Rightarrow \ln {\text{C}}x = \ln u \\
After restoring the substitution, we will get
\[ \Rightarrow \ln {\text{C}}x = \ln (\ln y)\]
Now taking exponential to base e, both sides, we will get
\Rightarrow {e^{\ln {\text{C}}x}} = {e^{\ln (\ln y)}} \\
\Rightarrow {\text{C}}x = \ln y \\
Again taking exponential to base e, we will get
\Rightarrow {e^{{\text{C}}x}} = {e^{\ln y}} \\
\Rightarrow {e^{{\text{C}}x}} = y \\
\Rightarrow y = {e^{{\text{C}}x}} \\
So \[y = {e^{{\text{C}}x}}\] is the general solution for the differential equation $\left( {y\ln y} \right)dx - xdy = 0$

Note: We have calculated the explicit general solution in this problem. Explicit solutions consist of dependent variables at the left hand side with power one and every other terms on the right hand side, whereas every solution which is not explicit is an implicit solution.