Section 4.5

#12. To verify that 2D-1 annihilates y=4ex/2, we apply 2D-1 TO 4ex/2. So (2D-1)4ex/2 = 2D(4ex/2) - 1(4ex/2) = 2(4ex/2/2) - 4ex/2 = 4ex/2 - 4ex/2 = 0. So it DOES annihilate.

#16. We need an annihilator for x3(1-5x), which expands to x3-5x4. An x3 would appear if there was a root of 0 four times, and x4 would appear if there was a root of 0 five times. So it takes (D-0)4 for the first and (D-0)5 for the second, which simplifies to D4 and D5. But we only need the bigger of those two, since it would also annihilate what the smaller one does. So our answer is D5.

#20. We must find an annihilator for 1+sin(x).
Since a 1 is actually 1e0x, the root that leads to 1 is m=0, which comes from a characteristic equation factor of (m-0), which comes from a differential operator of (D-0), or just D.
Since sin(x) comes from roots of 0ħ1i (remember, you must get both if you get one of them), then the characteristic equation must have (mħi) as factors, which comes from (Dħi) as the differential operators. More simply, (D+i)(D-i) = (D2+1).
Finally, the total annihilator must be the product of the pieces, so D(D2+1).

#38. The NH is: y ''' + 2y '' + y ' = 10. We first solve the corresponding LH: y ''' + 2y '' + y ' = 0. Using differential operators we get: (D3+2D2+D)y = 0, which factors to D(D+1)(D+1)y = 0. We can see that the characteristics are 0, -1 & -1, so the complementary solution is: yc = c1 + c2e-x + c3xe-x.
Now we look for yp. We notice that the RHS requires an annihilator of D, so we apply that to the entire NH, getting: D*D(D+1)(D+1)y = 0. The solution to this LH would be c1 + c2e-x + c3xe-x + c4x. We notice that the first three terms will be absorbed by the complementary solution, so that leaves the last term as the form for our particular solution. We use y = Ax as that form. Plugging in to the NH gives: 0 + 0 + A = 10, so A = 10. Thus yp = 10x.
Finally, our general solution is y = c1 + c2e-x + c3xe-x + 10x.

#44. The NH is: y '' + 2y ' + 2y = 5e6x. We first solve the corresponding LH: y '' + 2y ' + 2y = 0. Our characteristic equation is m2+2m+2m = 0, which solves via the quadratic formula to m = -1±i. The complementary solution is yc = c1e-xcos(x) + c2e-xsin(x).
Now we look for yp. We notice that the RHS requires an annihilator of D-6, so we apply that to the entire NH, getting: (D-6)(D2+2D+2)y = 0. The solution to this LH would be c1e-xcos(x) + c2e-xsin(x) + c3e6x . We notice that the first two terms will be absorbed by the complementary solution, so that leaves the last term as the form for our particular solution. We use y = Ae6x as that form. Plugging in to the NH gives: 36Ae6x + 12e6x + 2e6x = 5e6x, so 50A = 5, or A = 0.1 . Thus yp = 0.1e6x .
Finally, our general solution is y = c1e-xcos(x) + c2e-xsin(x) + 0.1e6x .

Return to the ODEs page