cdotter said:
Homework Statement
y''-3y'+2y=t e^{2t}+sin(5t)
Homework Equations
The Attempt at a Solution
I can get as far as this:
y_h(t)=c_1 e^{2t} + c_2 e^{t}
y_p(t)=At e^{2t} + B e^{2t} + C sin(5t) + D cos(5t)
My professor says we multiply At e^{2t} + B e^{2t} by t to give At^2 e^{2t} + B t e^{2t} because it's in the original equation. Why do we do this to the exp terms and not the Csin(5t) term, because that also shows up in the original equation?
The explanation is on the long side, so I'll try to give a shorter explanation using an example. Suppose your DE was y'' - 2y' = t.
The characteristic equation is r
2 - 2r = 0, or r(r - 2) = 0, or r = 0, or r = 2. An idependent pair of solutions to the homogeneous equation is y
1 = e
0, and y
2 = e
2t
Let's look at the diff. equation in terms of operators, where D means d/dt and D
2 means d
2/dt
2, and so on.
The original equation can be written as (D
2y - 2D)y = t, or (D
2 - 2D)y = t.
To get rid of the t on the right side, we can use the D
2 operator - the 2nd derivative of t is zero. Applying the D
2 operator to both sides, we get
D
2(D
2 - 2D)y = D
2t = 0.
The equation above is now fourth order, but is homogeneous.
The char. equation of this new equation is r
2(r
2 - 2r) = 0, or r
3(r - 2) = 0. The solutions are r = 0 (of multiplicity 3) and r = 2.
A set of linearly independent functions is {e
0, te
0, t
2e
0, e
2t} = {1, t, t
2, e
2t}.
The functions 1 and e
2t are solutions to the homogeneous form of the original equation (i.e., y'' - 2y' = 0). These two, plus the other two, t and t
2, are solutions to then new homogeneous fourth degree equation. What this means is that we should try t and t
2 for our particular solution of the nonhomogeneous equation y'' - 2y = t.
Notice that nothing happened with our other solution to the original homogenous equation, e
2t. The reason we added t and t
2 to our original solution 1, was because of repeated solutions r = 0 in our fourth order homogenous equation. Since there was no repetitition of the factor r - 2 in any of the characteristic equations, we don't have multiples by t of e
2t.
In your problem, a similar thing happens. r = 2 is a solution to the characteristic equation, and the addition of te
2t on the right side of the nonhomogeneous equation raises the multiplicity of the r - 2 factor.
Your equation could be written as (D
2 - 3D + 2)y = te
2t + cos(5t). In factored form on the left, this is (D - 1)(D - 2)y = te
2t + cos(5t).
The operator that annihilates the cos term on the right is D
2 + 25. The operator that annihilates the te
2t term is (D - 2)
2. The problem is that there is already a D - 2 term in the original, nonhomogeneous equation, so applying enough operators to annihilate the right side adds two more factors of D - 2, bringing the repetitions of this factor up to 3 in all. Adding a factor of D
2 + 25 added something that wasn't already there, so there's no change in the repetitition of existing factors in the operator equation, which BTW, should look an awful lot like the characteristic equation.
So, we are taking your 2nd order nonhomogenous equation and turning it into a sixth order homogeneous equation that looks like this: (D - 2)
3(D - 1)(D
2 + 25)y = 0. A basic set of solutions would be {e
2t, te
2t, t
2e
2t, e
t, cos(5t), sin(5t)}.
The functions e
2t and e
t are a basic set of functions for the solutions of the homogeneous form of your 2nd order equation. The other four are what you would use for your particular solution.
Hope that helps.
