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प्रश्न
Solve the following differential equation:-
\[x \log x\frac{dy}{dx} + y = \frac{2}{x}\log x\]
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उत्तर
We have,
\[x \log x\frac{dy}{dx} + y = \frac{2}{x}\log x\]
Dividing both sides by `x log x,` we get
\[\frac{dy}{dx} + \frac{y}{x \log x} = \frac{2}{x}\frac{\log x}{x \log x}\]
\[ \Rightarrow \frac{dy}{dx} + \frac{y}{x \log x} = \frac{2}{x^2}\]
\[ \Rightarrow \frac{dy}{dx} + \left( \frac{1}{x \log x} \right)y = \frac{2}{x^2}\]
\[\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get}\]
\[P = \frac{1}{x \log x} \]
\[Q = \frac{2}{x^2}\]
Now,
\[I . F . = e^{\int P\ dx} \]
\[ = e^{\int\frac{1}{x \log x}dx} \]
\[ = e^{\log\left| \left( \log x \right) \right|} \]
\[ = \log x\]
So, the solution is given by
\[y \times I . F . = \int Q \times I . F . dx + C\]
\[ \Rightarrow y\log x = 2\int\frac{1}{x^2} \times \log x dx + C\]
\[ \Rightarrow y\log x = I + C . . . . . . . . \left( 1 \right)\]
Where,
\[ \Rightarrow I = 2\log x\int\frac{1}{x^2} dx - 2\int\left[ \frac{d}{dx}\left( \log x \right)\int\frac{1}{x^2} dx \right]dx\]
\[ \Rightarrow I = \frac{- 2}{x}\log x + 2\int\left[ \frac{1}{x^2} \right]dx\]
\[ \Rightarrow I = \frac{- 2}{x}\log x - \frac{2}{x} . . . . . . . . \left( 2 \right)\]
From (1) and (2) we get
\[ \therefore y\log x = \frac{- 2}{x}\log x - \frac{2}{x} + C\]
\[ \Rightarrow y\log x = \frac{- 2}{x}\left( \log x + 1 \right) + C\]
