Question

d^2x/dt^2 + 6 dx/dt+ 8x = 5 sin 3t
differential equation using classical method. assume zero condition,anybodies can give ryt solution

Answer 1

Zero condition for both x and x'?

Answer 2

yes losser66

Answer 3

let wait for @tkhunny

Answer 4

did you ask him/her about this question?

Answer 5

Have you considered linear combinations of sin(3t) and cos(3t) as well as \(e^{2t}\) and \(e^{4t}\)?

Answer 6

Wasn't worth the wait, was it?

Answer 7

@tkhunny he asks for classical method which is laplace, am I right?

Answer 8

I don't think so. "Classical" is loosely associated with the Characteristic Equation and Undetermined Coefficients.

Answer 9

my teacher said that if if can't solve using clssical method try another one, it is laplace.

Answer 10

can we solve homogeneous part by undetermined and partial by laplace? (because of u(t) )

Answer 11

@tkhunny
if can't solve using that method.maybe there are any method that you know to solve these.

Answer 12

@jayven What is zero condition? :3

Answer 13

That makes no sense. Did you solve the characteristic equation to produce the solution of the homogeneous equation? If you had, you might have noticed my error. Those should be \(e^{-2t}\) and \(e^{-4t}\). I lost the negative sign somewhere along the way. Your first task is a simple algebra problem.

Answer 14

If you use Undetemined Coefficient method,
homogeneous part : CE: r^2 +6r +8 =0 gives r = -2 and r =-4
so, \(y_c = C_1e^{-2t}+C_2e^{-4t}\)
Partial solution forms the form of solution:
\(Y_p = Acos (3t) +Bsin(3t)\)
\(Y'_p = -3Asin (3t) +3Bcos(3t)\)
\(Y"_p = -9Acos (3t) -9Bsin(3t)\)
-----------------------------------
cos(3t)[-A+18B] + sin (3t)[-B-18A]= 5sin (3t)
---> -A +18 B = 0
-18A - B = 5
so A = -18/65
B = -1/65
general solution: \(y= C_1e^{-2t}+C_2e^{-4t}-\dfrac{18}{65}cos(3t)-\dfrac{1}{65}sin(3t)\)
Now, solve for condition: If you assume that y(0) =0
\(C_1+C_2 =\dfrac{18}{65}\)
and y'= \(-2C_1e^{-2t}-4C_2e^{-4t}+\dfrac{54}{65}sin(3t)+\dfrac{3}{65}cos(3t)\)
y'=0 --> \(-2C_1-4C_2=-\dfrac{3}{65}\)
solve for \(C_1~~,~~C_2\), you get \(C_1=69/130\) and \(C_2=-33/130\)
In conclusion,
\[y = \frac{69}{130}e^{-2t}-\frac{33}{130}e^{-4t}-\frac{18}{65}cos(3t)-\frac{1}{65}sin(3t)\]