Discussion
Solution of the differential equation \(x\frac{dy}{dx} + y = x^{4}\) with the boundary condition that \(y = 1\) at \(x = 1\) is :
Answer: D
The solution is \(y = \frac{x^{4}}{5} + \frac{4}{5x}\).
The left side is an exact derivative, \(x\frac{dy}{dx} + y = \frac{d}{dx}(xy)\), so
\[ \begin{aligned} \frac{d}{dx}(xy) &= x^{4} \\ xy &= \frac{x^{5}}{5} + C \\ y(1) = 1 &\Rightarrow C = \frac{4}{5} \\ y &= \frac{x^{4}}{5} + \frac{4}{5x} \end{aligned} \]
Checking the others: \(y = 5x^{4} - 4\) gives \(x y' + y = 25x^{4} - 4\), not \(x^{4}\); \(\frac{x^{4}}{5} + \frac{4x}{5}\) gives \(x^{4} + \frac{8x}{5}\); and \(\frac{4x^{4}}{5} + \frac{1}{5x}\) gives \(4x^{4}\), although it too satisfies \(y(1) = 1\).
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