No, while the difference represents the absolute magnitude of two numbers . . . for example . . .

The difference between 5 and 2 is . . . 3
The difference between 6.4 and 9.5 is . . . 3.1

. . . there is still the chance that the difference may be zero . . . in which case the difference is neither positive nor negative

. . . so in short . . . the answer is . . . <span>NO</span>

6 0

3 years ago

Solve the following ODE's: c) y* - 9y' + 18y = t^2

Nastasia [14]

Answer:

y = $C_1e^{3t}+C_2e^{6t}$ + $\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

Step-by-step explanation:

y''- 9 y' + 18 y = t²

solution of ordinary differential equation

using characteristics equation

m² - 9 m + 18 = 0

m² - 3 m - 6 m+ 18 = 0

(m-3)(m-6) = 0

m = 3,6

C.F. = $C_1e^{3t}+C_2e^{6t}$

now calculating P.I.

$P.I. = \frac{t^2}{D^2 - 9D +18}$

$P.I. = \dfrac{t^2}{(D-3)(D-6)}\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1-\frac{D}{6})^{-1}(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1+\frac{D}{6}+\frac{D^2}{36}+....)(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(1+\frac{D}{3}+\frac{D^2}{9}+....)(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

hence the complete solution

y = C.F. + P.I.

y = $C_1e^{3t}+C_2e^{6t}$ + $\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

7 0

2 years ago

PLEASE HELP !! ILL GIVE BRAINLIEST *EXTRA POINTS*.. IM GIVING 40 POINTS !! DONT SKIP :((.

PLEASE HELP !! ILL GIVE BRAINLIEST EXTRA POINTS.. IM GIVING 40 POINTS !! DONT SKIP :((.