Question

Solve the following differential equation:-

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

Answer 1

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

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

\[ \Rightarrow yx = \int x^3 + C\]

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