The general solution of the differential equation \(\displaystyle\frac{xdy}{dx}+ 4y = x^3, (x \ne 0)\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(y=\displaystyle\frac{x^3}{7}+cx^{-4}\) : c is arbitrary constant.
Dividing by x gives a linear differential equation \(\frac{dy}{dx} + Py = Q\):
\(\displaystyle \frac{dy}{dx} + \frac{4}{x}\,y = x^2\)
\(\displaystyle \text{I.F.} = e^{\int \frac{4}{x}\,dx} = e^{4\log|x|} = x^4\)
The solution is \(y \cdot \text{I.F.} = \int Q \cdot \text{I.F.}\,dx + c\):
\(\displaystyle x^4 y = \int x^6\,dx = \frac{x^7}{7} + c\)
Dividing by \(x^4\):
\(\displaystyle y = \frac{x^3}{7} + cx^{-4}\)