Question

Find one-parameter families of solution curves of the following differential equation:-

\[x\frac{dy}{dx} + y = x^4\]

Solve the following differential equation:-

\[x\frac{dy}{dx} + y = x^4\]

Answer 1

We have,
\[x\frac{dy}{dx} + y = x^4 \]
\[ \Rightarrow \frac{dy}{dx} + \frac{1}{x}y = x^3 . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dy}{dx} + Py = Q\]
where
\[P = \frac{1}{x} \]
\[Q = x^3 \]
\[ \therefore I . F . = e^{\int P\ dx} \]
\[ = e^{\int\frac{1}{x} dx} \]
\[ = e^{\log x} \]
\[ = x\]
\[\text{ Multiplying both sides of }\left( 1 \right)\text{ by }x, \text{ we get }\]
\[x \left( \frac{dy}{dx} + \frac{1}{x}y \right) = x . x^3 \]
\[ \Rightarrow x\frac{dy}{dx} + y = x^4 \]
Integrating both sides with respect to x, we get
\[xy = \int x^4 dx + C\]
\[ \Rightarrow xy = \frac{x^5}{5} + C\]
\[ \Rightarrow y = \frac{x^4}{5} + \frac{C}{x}\]
\[\text{ Hence, }y = \frac{x^4}{5} + \frac{C}{x}\text{ is the required solution.}\]