Question

\[\frac{dy}{dx} + \frac{y}{x} = x^3\]

Answer 1

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