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Given d3 ydx3 − d2 ydx2=2x + 3 The above equation can be written as (D3 − D2)y=2x + 3 Now the auxiliary equation is m3 − m2=0 which gives m=0, 0, 1 Hence the complimentary function is C.F.=C1 + C2x + c3ex Now we need to find the Particular Integral yp=1D2 + 3D − 10x(xx + 1)=1(D + 5)(D − 2)xex + 1 =ex1(D + 1 + 5)(D + 1 − 2)x + 1(D + 5)(D − 2)x =ex1(D + 6)(D − 1)x + 1(D + 5)(D − 2)x =ex1−6(1 + D6)(1 − D)x + 1−10(1 + D5)(1 − D2)x = −ex6(1 + D6)−1(1 − D)−1x − 110(1 − D5)−1(1 − D2)−1x = − ex6(1 + D6 + ⋯)(1 + D + D2)x − 110(1 + D5 ⋯)(1 + D2 + ⋯)x = − ex6(x + 1 + 16) − 110(x + 15 + 12) = − ex6(x + 76) − 110(x + 710) Hence the general solution is y=yc + yp=C1e−5x + C2e2x ± e6(x + 76) − 110(x + 710)
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