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