1. ## slightly unusual IVP

$u^{\prime\prime} + u = F(t), u(0)=0, u^{\prime}(0)=0$

$F(t)=\left\{\begin{array}{cc}F_0t, & 0 \leq t \leq \pi\\F_0(2\pi-t), & \pi < t\leq2\pi\\0, & 2\pi

Hint: treat each time interval separately, and match the solutions in the different intervals by requiring u and u' to be continuous functions of t.

I started this problem by trying to solve for the complimentary solution. The characteristic equation is r^2 + 1 =0. Meaning r = i.

Thus, I got:
$u(t) = c_1cos(t)+c_2sin(t)$

Using the initial conditions, I got c1 and c2 =0. I don't think this is right though...

2. Originally Posted by tautology
$u^{\prime\prime} + u = F(t), u(0)=0, u^{\prime}(0)=0$

$F(t)=\left\{\begin{array}{cc}F_0t, & 0 \leq t \leq \pi\\F_0(2\pi-t), & \pi < t\leq2\pi\\0, & 2\pi

Hint: treat each time interval separately, and match the solutions in the different intervals by requiring u and u' to be continuous functions of t.

I started this problem by trying to solve for the complimentary solution. The characteristic equation is r^2 + 1 =0. Meaning r = i.

Thus, I got:
$u(t) = c_1cos(t)+c_2sin(t)$

Using the initial conditions, I got c1 and c2 =0. I don't think this is right though...

CB

3. thanks for the link. I still don't get it though, I don't think we have covered stuff in that link. I will look more into it though.

4. Originally Posted by tautology
thanks for the link. I still don't get it though, I don't think we have covered stuff in that link. I will look more into it though.
You solve the IVP from $0$ to $\pi$, then use the solution and its derivative at $\pi$ as initial conditions to solve the IVP problem from $\pi$ to $2 \pi$ then use the values of the solution and its derivative at $2 \pi$ as initial conditions to solve the IVP for $t > 2\pi$.

CB

5. Originally Posted by tautology
$u^{\prime\prime} + u = F(t), u(0)=0, u^{\prime}(0)=0$

$F(t)=\left\{\begin{array}{cc}F_0t, & 0 \leq t \leq \pi\\F_0(2\pi-t), & \pi < t\leq2\pi\\0, & 2\pi

Hint: treat each time interval separately, and match the solutions in the different intervals by requiring u and u' to be continuous functions of t.

I started this problem by trying to solve for the complimentary solution. The characteristic equation is r^2 + 1 =0. Meaning r = i.

Thus, I got:
$u(t) = c_1cos(t)+c_2sin(t)$

Using the initial conditions, I got c1 and c2 =0. I don't think this is right though...
Put the initial values into the entire solution, not just the solution to the associated homogenous equation (your "complementary" solution), to determine the constants.

First solve $u"+ u= F_0t$. You should get $u(t)= c_1cos(t)+ c_2sin(t)+ F_0t$. Putting u(0)= 0, u'(0)= 0 into that gives you $u(0)= c_1= 0$ and $u'(0)= c_2+ F_0= 0$ so while $c_1= 0$, $c_2= -F_0$ The solution to the entire equation is $u(t)= -F_0 sin(t)+ F_0t$ for [tex]0\le t\le \pi[/itex]. Since $sin(\pi)= 0$ and $cos(\pi)= -1$, at $t= \pi$ we have $u(\pi)= F_0\pi$ and $u'(\pi)= 2F_0$.

Now solve $u"+ u= 2\pi F_0- F_0t$ with initial conditions $u(\pi)= F_0\pi$ and $u'(\pi)= 2F_0$. That will give you u for $\pi\le t\le 2\pi$. Determine $u(2\pi)$ and $u'(2\pi)$ from that.

Finally, solve $u"+ u= 0$ with initial conditions [tex]u(2\pi)[/math ] and $u'(2\pi)$ equal to the values you just got.

6. The problem is very similar to this...

http://www.mathhelpforum.com/math-he...-function.html

Indicating with $F(s)$ the LT of $f(t)$ the DE is written in terms of the complex variable $s$ as...

$s^{2}\cdot Y(s) - s\cdot y(0) - y^{'} (0) + Y(s) = F(s)$ (1)

Since is $y(0) = y^{'} (0) =0$ the solution is the 'only' $y_{forced} (t)$ , and because is...

$F(s)= \frac{f_{0}\cdot(1-e^{-\pi s})^{2}}{s^{2}}$ (2)

... is...

$Y(s)= \frac{f_{0}\cdot (1-e^{-\pi s})^{2}}{s^{2}\cdot (1+s^{2})}$ (3)

The solution is then...

$y(t) = \mathcal{L}^{-1} \{Y(s)\}$ (4)

Kind regards

$\chi$ $\sigma$