To solve the differential equation y(4) 2y(3) 3y′′ 2y′ y = 0, we can use the method of characteristic roots.
First, let's assume that the solution is of the form y = e^(rt). Then, we can differentiate y with respect to t to get y′ = re^(rt), differentiate again to get y′′ = r^2e^(rt), and differentiate one more time to get y(4) = r^4e^(rt).
Substituting these expressions into the original differential equation, we get:
r^4e^(rt) - 2r^3e^(rt) + 3r^2e^(rt) - 2re^(rt) + e^(rt) = 0
Dividing both sides by e^(rt), we can simplify the equation to:
r^4 - 2r^3 + 3r^2 - 2r + 1 = 0
This is a fourth-order polynomial equation that we can solve using the characteristic roots method.
Expanding (r2 r +1)2, we get:
r^4 + 2r^3 + 3r^2 + 2r + 1 = 0
Comparing this to the polynomial equation we obtained earlier, we can see that they are identical except for the sign of the middle term. Therefore, the characteristic roots of our differential equation are the roots of (r2 r +1)2, which are:
r = -1 (double root) and r = -i (double complex root)
This means that the general solution of the differential equation is:
y = c1e^(-t) + c2te^(-t) + c3cos(t) + c4sin(t)
where c1, c2, c3, and c4 are arbitrary constants that can be determined from initial or boundary conditions.