Question

Solve the differential equation: $x\dfrac{dy}{dx}=y\left( \log y-\log x+1 \right)$.
(a) $\log y-\log x=x+c$
(b) $\log y+\log x=cx$
(c) $\log y-\log x=cx$
(d) $\log y-\log x+cx=0$

Answer 1

Hint: Divide by ‘x’ both sides and use the property of logarithm give by: $\log m-\log n=\left( \log \dfrac{m}{n} \right)$, to change the logarithmic term into $\log \left( \dfrac{y}{x} \right)$. Substitute, $y=vx$, then find $\dfrac{dy}{dx}$ in terms of v and x. Now, separate terms containing ‘x’ one side and the terms containing ‘v’ another side. Integrate both sides with respect to the variable x and v respectively. Finally substitute the value of ‘v’ and simply the expression.

Complete step-by-step solution -
We have been given the differential equation: $x\dfrac{dy}{dx}=y\left( \log y-\log x+1 \right)$.
Dividing both sides by ‘x’, we get,
$\dfrac{dy}{dx}=\dfrac{y\left( \log y-\log x+1 \right)}{x}$
This can be written as,
$\dfrac{dy}{dx}=\dfrac{y}{x}\times \left( \log y-\log x+1 \right)$
Now, using the property of log given by: $\log m-\log n=\left( \log \dfrac{m}{n} \right)$, we get,
$\dfrac{dy}{dx}=\dfrac{y}{x}\times \left( \log \dfrac{y}{x}+1 \right)$
Substituting, $y=vx$, we have,
$\dfrac{dy}{dx}=v\left( \log v+1 \right)......................(i)$
Now, y = vx. Therefore, differentiating both sides with respect to ‘x’ using chain rule, we get,
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{dvx}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}=v\dfrac{dx}{dx}+x\dfrac{dv}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}.............................(ii) \\
\end{align}$
Substituting the value of $\dfrac{dy}{dx}$ from equation (ii) in equation (i), we get,
$\begin{align}
  & v+x\dfrac{dv}{dx}=v\left( \log v+1 \right) \\
 & \Rightarrow v+x\dfrac{dv}{dx}=v\log v+v \\
\end{align}$
Cancelling ‘v’ from both sides, we get,
$x\dfrac{dv}{dx}=v\log v$
Now, separating the variables containing ‘v’ to the L.H.S and the variables containing ‘x’ to the R.H.S, we have,
\[\dfrac{dv}{v\log v}=\dfrac{dx}{x}\]
Integrating the above expression, we get,
\[\int{\dfrac{dv}{v\log v}=\int{\dfrac{dx}{x}}}\]
Let us find the integral of L.H.S first.
Assuming, $\log v=m$, we have,
$\begin{align}
  & d\log v=dm \\
 & \Rightarrow \dfrac{1}{v}dv=dm \\
\end{align}$
Substituting these value in the L.H.S integral, we get,
$\int{\dfrac{dm}{m}=}\int{\dfrac{dx}{x}}$
We know that, $\int{\dfrac{dx}{x}=}\log x$. Therefore,
$\log m=\log x+\log c$, where $\log c$ is any arbitrary constant.
Substituting the value of ‘m’ we get,
$\log \left( \log v \right)=\log x+\log c$
Now, substituting the value of ‘v’ we get,
$\log \left( \log \dfrac{y}{x} \right)=\log x+\log c$
Again using the properties of logarithm given by:
$\log m+\log n=\log mn\text{ and }\log m-\log n=\log \dfrac{m}{n}$, we have,
$\log \left( \log y-\log x \right)=\log cx$
Removing log from both sides, we get,
$\log y-\log x=cx$
Hence, option (c) is the correct answer.

Note: One may note that, we have chosen ‘$\log c$’ and not ‘c’ as our arbitrary constant or you can say constant of integration. The reason is that we can easily use the properties of logarithm and simplify the expression. If we will use ‘c’ as the arbitrary constant then we have to use the property of exponents to simplify which will be lengthy. However, the answer will not change. The above method of finding a solution is a mixture of substitution method and variable separable method.