परिसीमा शर्तों y(0) = y(π) = 0 के साथ अवकल समीकरण \(y^{\prime \prime}+\frac{y}{4}=\frac{x}{2} \), जहां 0 ≤ x ≤ π, का समाधान y(x) है
1
\(\frac{2}{\pi} \sum_{n=1}^{\infty}(-1)^n \frac{\pi}{n} \frac{\sin n x}{\frac{1}{4}-n^2} \)
2
\(\frac{2}{\pi} \sum_{n=1}^{\infty}(-1)^n \frac{\pi}{2 n} \frac{\sin n x}{\frac{1}{4} n^2} \)
3
\(\frac{2}{\pi} \sum_{n=1}^{\infty}(-1)^{n+1} \frac{\pi}{n} \frac{\sin n x}{\frac{1}{4} n^2} \)
4
\(\frac{2}{\pi} \sum_{n=1}^{\infty}(-1)^{n+1} \frac{\pi}{2 n} \frac{\sin n x}{\frac{1}{4}-n^2} \)