The characteristic or auxiliary equation for 2y''+3y'+y = 0 is 2λ2+3λ+1 = (λ+1)(2λ+1) = 0, so λ1 = -1 & λ2 = -1/2; these roots are real and distinct. Therefore, the general solution to the associated homogeneous differential equation is yh = C1e-t+C2e-t/2.
Next, solve 2y''+3y'+y = t2. Assume a particular solution of the form yp =A2t2+A1t+A0 and compare to yh = C1e-t+C2e-t/2; yp and yh have no terms in common, except for a possible multiplicative constant. There is then no need to modify yp.
Place yp in 2y''+3y'+y = t2 to obtain 2(2A2)+3(2A2t+A1)+(A2t2+A1t+A0) = t2. Rewrite as A2t2+(6A2+A1)t+(4A2+3A1+A0) = t2. This gives A2 = 1 & 6(1)+A1 =0 or A1 = -6. Also, 4(1) + 3(-6) +A0 = 0 or A0 = 14.
Then yp = t2-6t+14 and the general solution is y = yh+yp = C1e-t+C2e-t/2+t2-6t+14.
Next, consider 2y''+3y'+y = 3sin t. Again, as above, yh = C1e-t+C2e-t/2. Assume a particular solution of the form yp = A0sin t+B0cos t, which needs no modification. Putting yp into 2y''+3y'+y = 3sin t, write 2(-A0sin t-B0cos t)+3(A0cos t-B0sin t)+A0sin t+B0cos t = 3sin t. Rewrite as (-2A0-3B0+A0)sin t+(-2B0+3A0+B0)cos t = 3sin t. This gives {-A0-3B0 = 3; 3A0-B0 = 0} or A0 = -3/10 & B0 = -9/10.
Then yp = (-3/10)sin t-(9/10)cos t. With the particular solution of 2y''+3y'+y = t2 being t2-6t+14 and the particular solution of 2y''+3y'+y = 3sin t being (-3/10)sin t-(9/10)cos t, a particular solution to 2y''+3y'+y = t2+3sin t is then t2-6t+14-(3/10)sin t-(9/10)cos t. Finally, when combined with the complementary solution of yh = C1e-t+C2e-t/2, the general solution sought is written as y = C1e-t+C2e-t/2+t2-6t+14-(3/10)sin t-(9/10)cos t.
The answer supplied by Janielly V. as a check in the statement of the problem differs only by the operation sign just before 6t; it should be a "minus" instead of a "plus".
