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D^2y +4y = sin^2 x |
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Answer» Symbolic form of the given equation, (D2+4)y = sin 2x Corresponding auxiliary equation, D2+4 = 0 i.e. D = ±2i Thus, y(C.F.) = (c1 cos 2x + c2 sin 2x) y(P.I.) \(=\frac1{D^2+4}\)sin 2x; Replace D2 = -a2 = -4, but here f(-a2) = 0 [In case of failure, \(\frac1{f(D^2)}\)sin(ax+b) \(=x\frac1{f'(-a^2)}\)sin(ax+b)] Implying y(P.I.) \(=x\frac{1}{2.D}sin\,2x=\frac{x}{2}(\frac{-cos\,2x}{2})=-\frac{x\,cos\,2x}{4}.\) Hence the complete solution, y = (c1 cos 2x + c2 sin 2x) \(-\frac{x\,cos\,2x}4.\) |
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