Let us solve the differential equation (D2−2D+3)y=x2−1. For this let us solve the characteristic equation of the homogeneous equation (D2−2D+3)y=0:
k2−2k+3=0
(k−1)2+2=0
(k−1)2=−2
k1=1+i2 and k2=1−i2.
It follows that the general solution of the equation (D2−2D+3)y=x2−1 is of the form y=ex(C1cos(2x)+C2sin(2x))+yp, where yp=Ax2+Bx+C. It follows that yp′=2Ax+B and yp′′=2A. Therefore, 2A−2(2Ax+B)+3(Ax2+Bx+C)=x2−1. We conclude that 3Ax2+(3B−4A)x+(2A−2B+3C)=x2−1. Consequently, 3A=1,3B−4A=0,2A−2B+3C=−1. It follows that A=31,B=94,C=31(−1−32+98)=−277.
We conclude that the general solution of the differential equation (D2−2D+3)y=x2−1 is
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