Question

Solve the following differential equation:-

\[x\frac{dy}{dx} + 2y = x^2 , x \neq 0\]

Answer 1

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

\[y \times I . F . = \int Q \times I . F . dx + C\]

\[ \Rightarrow y x^2 = \int x^3 dx + C\]

\[ \Rightarrow y x^2 = \frac{x^4}{4} + C\]

\[ \Rightarrow y = \frac{x^2}{4} + C x^{- 2}\]