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Question
Solve the following differential equation:-
\[x\frac{dy}{dx} + 2y = x^2 , x \neq 0\]
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Solution
We have,
\[x\frac{dy}{dx} + 2y = x^2 \]
\[ \Rightarrow \frac{dy}{dx} + \frac{2}{x}y = x\]
\[\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get}\]
\[P = \frac{2}{x} \]
\[Q = x\]
Now,
\[I . F . = e^{2\int\frac{1}{x}dx} \]
\[ = e^{2\log \left| x \right|} \]
\[ = x^2 \]
So, the solution is given by
\[y \times I . F . = \int Q \times I . F . dx + C\]
\[ \Rightarrow y x^2 = \int x^3 dx + C\]
\[ \Rightarrow y x^2 = \frac{x^4}{4} + C\]
\[ \Rightarrow y = \frac{x^2}{4} + C x^{- 2}\]
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