What is the general solution of the differential equation y''' - 3y'-2y =0 ?
2 Answers
y = c_1e^(-x) + c_2xe^(-x) + c_3e^(2x)
Well, this is a linear third-order ordinary differential equation. We can assume a solution of
y = e^(rx)
so that
r^3 e^(rx) - 3re^(rx) - 2e^(rx) = 0
And since
r^3 - 3r - 2 = 0
Now we need to figure out how this can be factored. By synthetic division, we have a guess for
ul(-1)|" "1" "" "0" "-3" "-2
+" "" "" "" "-1" "" "1" "" "2
"----------------------------------------"
" "" "" "1" "-1" "-2" "" "0
And we happen to be correct. Thus, we obtain a factorization of:
(r+1)(r^2 - r - 2) = 0
(r+1)^2(r-2) = 0
So, the roots are
color(blue)(y = c_1e^(-x) + c_2xe^(-x) + c_3e^(2x)) Note that the multiplicity-2 root has two unique constants in its part of the linear combination, but the second root instance is distinguished by another
x term in front.
And we should check whether we are correct or not... Do we still have that
y''' - 3y' - 2y = 0 ?
Firstly, we have a straightforward differentiation using the product rule for the first term:
y''' = -c_1e^(-x) + c_2(d^2)/(dx^2)[-xe^(-x) + e^(-x)] + 8c_3e^(2x)
= -c_1e^(-x) + c_2(d)/(dx)[-[-xe^(-x) + e^(-x)] - e^(-x)] + 8c_3e^(2x)
= -c_1e^(-x) + c_2(d)/(dx)[xe^(-x) - 2e^(-x)] + 8c_3e^(2x)
= -c_1e^(-x) + c_2[[-xe^(-x) + e^(-x)] + 2e^(-x)] + 8c_3e^(2x)
= -c_1e^(-x) - c_2[xe^(-x) - 3e^(-x)] + 8c_3e^(2x)
And for the second term:
y' = -c_1e^(-x) - c_2[xe^(-x) - e^(-x)] + 2c_3e^(2x)
So, we shall verify whether our solution works.
y''' - 3y' - 2y stackrel(?" ")(=) 0
= {-c_1e^(-x) - c_2[xe^(-x) - 3e^(-x)] + 8c_3e^(2x)} - 3{-c_1e^(-x) - c_2[xe^(-x) - e^(-x)] + 2c_3e^(2x)} - 2{c_1e^(-x) + c_2xe^(-x) + c_3e^(2x)}
= -c_1e^(-x) + c_2xe^(-x) - 3c_2e^(-x) + 8c_3e^(2x) + 3c_1e^(-x) + 3c_2xe^(-x) - 3c_2e^(-x) - 6c_3e^(2x) - 2c_1e^(-x) - 2c_2xe^(-x) - 2c_3e^(2x)
= -c_1e^(-x) - c_2xe^(-x) + cancel(3c_2e^(-x)) + cancel(8c_3e^(2x)) + 3c_1e^(-x) + 3c_2xe^(-x) - cancel(3c_2e^(-x)) - cancel(6c_3e^(2x)) - 2c_1e^(-x) - 2c_2xe^(-x) - cancel(2c_3e^(2x))
= -c_1e^(-x) - cancel(c_2xe^(-x)) + 3c_1e^(-x) + cancel(3c_2xe^(-x)) - 2c_1e^(-x) - cancel(2c_2xe^(-x))
= -cancel(c_1e^(-x)) + cancel(3c_1e^(-x)) - cancel(2c_1e^(-x))
= 0 color(blue)(sqrt"")
So indeed, we do have the general solution.
y = Axe^(-x) + Be^(-x) + Ce^(2x)
Explanation:
We have:
y''' - 3y'-2y =0 ..... [A]
This is a Third order linear Homogeneous Differentiation Equation with constant coefficients. The standard approach is to find a solution,
Complementary Function
The Auxiliary equation associated with the homogeneous equation of [A] is:
m^3 -3m-2 = 0
The challenge with higher order Differential Equation is solving the associated higher order Auxiliary equation. By inspection we see
(m+1)(m^2-m-2) = 0
:. (m+1)(m+1)(m-2) = 0
:. (m+1)^2(m-2) = 0
Which has three real solutions, one distinct and one repeated twice:
m=-1,-1,2 .
The roots of the auxiliary equation determine parts of the solution, which if linearly independent then the superposition of the solutions form the full general solution.
- Real distinct roots
m=alpha,beta, ... will yield linearly independent solutions of the formy_1=Ae^(alphax) ,y_2=Be^(betax) , ... - Real repeated roots
m=alpha , will yield a solution of the formy=(Ax+B)e^(alphax) where the polynomial has the same degree as the repeat. - Complex roots (which must occur as conjugate pairs)
m=p+-qi will yield a pairs linearly independent solutions of the formy=e^(px)(Acos(qx)+Bsin(qx))
Thus the solution of the homogeneous equation is:
y = (Ax+B)e^(-1x) + Ce^(2x)
\ \ = Axe^(-x) + Be^(-x) + Ce^(2x)
Note this solution has