Free Version
Moderate

# IVP: Triangular Forcing

DIFFEQ-E3OLHN

Let $u(t)$ be the Heaviside step function (some people use the notation $H(t)$ for it). Consider a 'triangle forcing' $g(t)=t(u(t)-u(t-1))$, which equals $t$ on the interval $[0,1]$ and zero everywhere else.

The solution to $y''+y=g(t)$, $y(0)=y'(0)=0$ is:

A

$t-\sin(t)-u_1(t)(t-1-\sin(t-1))$

B

$h(t)-u_1(t)h(t-1)+(u_1(t)-u_1(t)\cos(t-1))$ where $h(t)=t-\sin(t)$

C

$h(t)-u_1(t)h(t-1)+(u_1(t)-u_1(t)\cos t)$ where $h(t)=t-\sin(t)$

D

$h(t)-u_1(t)h(t-1)+(-u_1(t)+u_1(t)\cos(t-1))$ where $h(t)=t-\sin(t)$