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1.	(D^2+4)y=sin^2+x^4
Answer» We have: (D² + 4)y = sin2x + x⁴ C.F:- m² + 4 = 0 m = \(\pm\)2i y = c₁cos2x + c₂ sin2x P.I. y = \(\frac{1}{D^2+4}\)(sin2x + x⁴) = \(\frac{1}{D^2+4}\)(sin2x) + \(\frac{1}{D^2+4}\)x⁴ = \(\frac{-\mathrm{x}}{2\times2}\)cos2x + \(\frac{1}{4\big(1+\frac{D^2}{4}\big)}\) x⁴ ⁼\(\frac{-\mathrm{x}}{4}\) cos2x + \(\frac{1}{4}\)\(\big(1-\frac{D^2}{4}+\frac{D^4}{16}-\frac{D^8}{64}+...\big)\)x⁴ ⁼\(\frac{-\mathrm{x}}{4}\) cos2x + \(\frac{\mathrm{x}^4}{4}\) - \(\frac{D^2\mathrm{x}^4}{4}\) + \(\frac{D^4\mathrm{x}^4}{16}\) - 0 + 0 + ...... = \(\frac{-\mathrm{x}}{4}\) cos2x + \(\frac{\mathrm{x}^4}{4}\) - \(\frac{12\mathrm{x}^2}{4}\) + \(\frac{24}{16}\) complete solution: y = C.F + P.I = (c₁\(\frac{-\mathrm{x}}{4}\))cos 2x + C₂ sin2x + \(\frac{\mathrm{x}^4}{4}\) - 3x² + \(\frac{3}{2}\)

(D^2+4)y=sin^2+x^4

Answer»

We have:

(D² + 4)y = sin2x + x⁴

C.F:- m² + 4 = 0

m = \(\pm\)2i

y = c₁cos2x + c₂ sin2x

P.I. y = \(\frac{1}{D^2+4}\)(sin2x + x⁴)

= \(\frac{1}{D^2+4}\)(sin2x) + \(\frac{1}{D^2+4}\)x⁴

= \(\frac{-\mathrm{x}}{2\times2}\)cos2x + \(\frac{1}{4\big(1+\frac{D^2}{4}\big)}\) x⁴

⁼\(\frac{-\mathrm{x}}{4}\) cos2x + \(\frac{1}{4}\)\(\big(1-\frac{D^2}{4}+\frac{D^4}{16}-\frac{D^8}{64}+...\big)\)x⁴

⁼\(\frac{-\mathrm{x}}{4}\) cos2x + \(\frac{\mathrm{x}^4}{4}\) - \(\frac{D^2\mathrm{x}^4}{4}\) + \(\frac{D^4\mathrm{x}^4}{16}\) - 0 + 0 + ......

= \(\frac{-\mathrm{x}}{4}\) cos2x + \(\frac{\mathrm{x}^4}{4}\) - \(\frac{12\mathrm{x}^2}{4}\) + \(\frac{24}{16}\)

complete solution:

y = C.F + P.I = (c₁\(\frac{-\mathrm{x}}{4}\))cos 2x + C₂ sin2x + \(\frac{\mathrm{x}^4}{4}\) - 3x² + \(\frac{3}{2}\)

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