Question

x dy = (2y + 2x⁴ + x²) dx

Answer 1

We have, 
\[x dy = \left( 2y + 2 x^4 + x^2 \right)dx\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2}{x}y + 2 x^3 + x\]
\[ \Rightarrow \frac{dy}{dx} - \frac{2}{x}y = 2 x^3 + x . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form 
\[\frac{dy}{dx} + Py = Q\]
where
\[P = - \frac{2}{x}\]
\[Q = 2 x^3 + x\]
\[ \therefore I.F. = e^{\int P\ dx} \]
\[ = e^{- \int\frac{2}{x} dx} \]
\[ = e^{- 2\log x} \]
\[ = \frac{1}{x^2}\]
\[\text{ Multiplying both sides of } \left( 1 \right)\text{ by }\frac{1}{x^2}, \text{ we get }\]
\[\frac{1}{x^2} \left( \frac{dy}{dx} - \frac{2}{x}y \right) = \frac{1}{x^2} \left( 2 x^3 + x \right)\]
\[ \Rightarrow \frac{1}{x^2}\frac{dy}{dx} - \frac{2}{x^3}y = 2x + \frac{1}{x}\]
Integrating both sides with respect to x, we get
\[\frac{1}{x^2}y = \int\left( 2x + \frac{1}{x} \right)dx + C\]
\[ \Rightarrow \frac{1}{x^2}y = x^2 + \log x + C\]
\[ \Rightarrow y = x^4 + x^2 \log x + C x^2 \]
\[\text{Hence, }y = x^4 + x^2 \log x + C x^2\text{ is the required solution.}\]