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The particular integral value of (D^{2}+4)y=sin^{2}x is |
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Answer» P.I. = \(\frac{1}{D^2+4}\) sin2x = \(\frac{1}{D^2+4}\) \(\Big(\frac{1-cos2x}{2}\Big)\) = \(\frac{1}{2}\) \(\frac{1}{D^2+4}\) ex - \(\frac{1}{2}\) x \(\frac{1}{D^2+4}\) cos2x = \(\frac{1}{8}-\frac{1}{2}\times \frac{x}{2x^2}\) sin2x = \(\frac{1}{8}-\frac{x}{8}\) sin2x = \(\frac{1}{8}\) (1-sin2x) |
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