Question

Suppose you are solving a linear homogenous differential equation. State the general solution for each of the following situations. You may assume that the dependent variable is y, and y = y(t). a. Second-order constant-coefficient equation, with auxiliary equation roots r = -2 +V-49 b. Second-order constant-coefficient equation, with auxiliary equation roots r = -2 + 149 c. Second-order Cauchy-Euler equation, with auxiliary equation roots r = -2 EVO d. Fourth-order constant-coefficient equation, with auxiliary equation roots r = -6, -6, +9i

Answer 1

(a) The roots of the auxiliary equation are complex, given by $-2\pm \sqrt{-49}=-2\pm 7\sqrt{-1}=-2\pm7i$ , so the general solution is given by

y(t) = e-2t (c1 cos 7x + C2 sin 7x)

(b) This time, the roots of the auxiliary equation are real and distinct, given by , so the general solution is given by

-2 + 49 = -2+ 7 = 5.-9

$\LARGE y(t)=c_1e^{5t}+c_2e^{-9t}$

(c) The auxiliary equation of the second order Cauchy Euler equation has two real and equal roots given by $-2\pm \sqrt 0=-2,-2$ , so the general solution is given by

$\LARGE y(t)=c_1t^{-2}\ln(t)+c_2t^{-2}=c_1 \frac{\ln(t)}{t^2}+c_2\frac{1}{t^2}$

(d) The auxiliary equation has two repeated real roots and one pair of complex conjugate roots given by so the general solution is given by

$\LARGE y(t)=(c_1+c_2t)e^{-6t}+c_3\cos 9t +c_4 \sin 9t$
(the e^t term on the right becomes one as the complex roots have no real part)