- #1
musemonkey
- 25
- 0
1. This is exercise 3.19.15 from Boyce & DiPrima's Differential Equations.
[tex] u'' + u = F(t), u(0) = 0, u'(0) = 0 [/tex]
[tex] 0 \leq t < \pi , F(t) = F_0t [/tex]
[tex] \pi \leq t \leq 2\pi , F(t) = F_0(2\pi-t) [/tex]
[tex] t > 2\pi , F(t) = 0 [/tex]
Solve for [tex] u(t) [/tex].
2. The solution to the homogenous equation is
[tex] u'' + u = 0 [/tex]
is [tex] u(t) = c_1 cos(t) + c_2 sin(t) [/tex].
3. Plugging in the boundary conditions yields
[tex] u(0) = c_1 + 0 = 0 [/tex]
[tex] u'(t) = - c_1 sin(t) + c_2 cos(t) [/tex]
[tex] u'(0) = c_2 = 0 [/tex].
Then a particular solution of the form [tex] u_p = A t [/tex] works
[tex] u'' + u = At = F_0t [/tex],
and so the solution for [tex] 0 \leq t < \pi [/tex] I find to be [tex] u_1(t) = F_0 t [/tex]. But the solution in the back of the textbook states [tex] u_1(t) = F_0t - F_0
sin(t) [/tex], which clearly also satisfies the differential equation and the initial value conditions, but it's inconsistent with what I get. Where did I go wrong?
[tex] u'' + u = F(t), u(0) = 0, u'(0) = 0 [/tex]
[tex] 0 \leq t < \pi , F(t) = F_0t [/tex]
[tex] \pi \leq t \leq 2\pi , F(t) = F_0(2\pi-t) [/tex]
[tex] t > 2\pi , F(t) = 0 [/tex]
Solve for [tex] u(t) [/tex].
2. The solution to the homogenous equation is
[tex] u'' + u = 0 [/tex]
is [tex] u(t) = c_1 cos(t) + c_2 sin(t) [/tex].
3. Plugging in the boundary conditions yields
[tex] u(0) = c_1 + 0 = 0 [/tex]
[tex] u'(t) = - c_1 sin(t) + c_2 cos(t) [/tex]
[tex] u'(0) = c_2 = 0 [/tex].
Then a particular solution of the form [tex] u_p = A t [/tex] works
[tex] u'' + u = At = F_0t [/tex],
and so the solution for [tex] 0 \leq t < \pi [/tex] I find to be [tex] u_1(t) = F_0 t [/tex]. But the solution in the back of the textbook states [tex] u_1(t) = F_0t - F_0
sin(t) [/tex], which clearly also satisfies the differential equation and the initial value conditions, but it's inconsistent with what I get. Where did I go wrong?