Please answer the follwing questions Consider the homogeneous equation y\'\" + 8
ID: 3342056 • Letter: P
Question
Please answer the follwing questions
Consider the homogeneous equation y'" + 8y = 0 Write clown the characteristic equation and solve for r (there will be 3 solutions). For r1, r2, and r3 write the corresponding solutions y1, y2, and y3 to the DE just as you did in chapter 3. Show that the solutions y1, y2, and y3 are linearly independent (i.e., different) by taking the Wronskian W(y1,y2,y3). The general solution of the DE is y = c1y1 + c2y2 + C3y3. Use the method of problem 1 to find the general solution of the differential equation y(4) + 6y'" + 12y" + 10 y' + 3 y = 0. Write the general form of the solution to the DE y(4) + 6y'" + 12y" + 10 y' + 3 y = 3e-t + e-3t + 2 cos t. Do not solve for the unknown coefficients.Explanation / Answer
(D^3+8) = (D+2)(D^2-2D+4)
r = -2, 1 + sqrt(3)i, 1 - sqrt(3)i
y1 = e^-2x
y2=e^(1+sqrt(3)i)x
y3=e^(1-sqrt(3)i)x
The Wronskian is
e^-2x e^(1+sqrt(3)i)x e^(1-sqrt(3)i)x
-2e^-2x (1+sqrt(3)i)e^(1+sqrt(3)i)x (1-sqrt(3)i)e^(1-sqrt(3)i)x
4e^-2x (1+sqrt(3)i)^2e^(1+sqrt(3)i)x (1-sqrt(3)i)^2e^(1-sqrt(3)i)x
Then, the determinant is
e^-2x e^(1+sqrt(3)i)x e^(1-sqrt(3)i)x
((1+sqrt(3)i) (1-sqrt(3)i)^2 - (1-sqrt(3)i) (1+sqrt(3)i)^2
-(-2(1-sqrt(3)i)^2-4(1-sqrt(3)i)
+ (-2(1+sqrt(3)i)^2-4(1+sqrt(3)i)) =
1
(4 - 4 sqrt(3)i -(4 + 4 sqrt(3)i)
+(-4-4sqrt(3)i + 4 - 4sqrt(3)i)
+(4-4sqrt(3)i - 4- 4 sqrt(3)i) =
-24 sqrt(3) i is not equal to 0
Independence
The general solution is either c1e^-2x + c2e^(1+sqrt(3)i)x+c3e^(1-sqrt(3)i)x or
c1e^-2x + c2e^xcos(sqrt(3)x) + c3e^xsin(sqrt(3)x)
2)D^4 + 6D^3 + 12D^2 +10D + 3 = 0
Solve for a where
(D^2+aD+1)(D^2+(6-a)D+3) = 0
(we have 6-a because the cubic term is 6)
3a +6-a = 10 for the linear term
2a = 4
a = 2
(D^2+2D+1)(D^2+4D+3) = 0
(D+1)^3(D+3) = 0
We have a multiple of 3 for -1 and -3
Then, the solution is
c1e^-3x + c2e^-x + c3xe^-x + c4x^2e^-x
3)
On this problem, I see that we need to use t rather than x.
This is the same equation as in 2 on the left hand side but with a non-0 right-hand side.
Note that 2 of the terms on the right hand side correspond to roots for the homogeneous solution.
Thus, the solution is
c1e^-3t + c2e^-t+ c3te^-t + c4t^2e^-t + d1t^3e^-t + d2te^-3t + d3cost + d4sint