Question

Find one-parameter families of solution curves of the following differential equation:-

\[x \log x\frac{dy}{dx} + y = 2 \log x\]

Solve the following differential equation:-

\[x \log x\frac{dy}{dx} + y = 2 \log x\]

Answer 1

We have, 
\[x \log x\frac{dy}{dx} + y = 2\log x\]
Dividing both sides by x \log x, we get
\[\frac{dy}{dx} + \frac{y}{x \log x} = 2\frac{\log x}{x \log x}\]
\[ \Rightarrow \frac{dy}{dx} + \frac{y}{x \log x} = \frac{2}{x}\]
\[ \Rightarrow \frac{dy}{dx} + \left( \frac{1}{x \log x} \right)y = \frac{2}{x}\]
\[\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get }\]
\[P = \frac{1}{x \log x}\]
\[Q = \frac{2}{x}\]
Now,
\[I.F. = e^{\int P\ dx} = e^{\int\frac{1}{x \log x}dx} \]
\[ = e^{log\left| \log x \right|} \]
\[ = \log x\]
So, the solution is given by
\[y \times I . F . = \int Q \times I . F . dx + C\]
\[ \Rightarrow y \log x = 2\int\frac{1}{x} \times \log x dx + C\]
\[\text{Putting }\log x = t\]
\[ \Rightarrow \frac{1}{x}dx = dt\]
\[ \therefore y \log x = 2\int t\ dt + C\]
\[ \Rightarrow y \log x = \frac{2 t^2}{2} + C\]
\[ \Rightarrow y \log x = t^2 + C\]
\[ \Rightarrow y \log x = \left( \log x \right)^2 + C .............\left( \because \log x = t \right)\]
\[ \Rightarrow y = \log x + \frac{C}{\log x}\]