1. Dif EQ T/F

State whether the following are true or false. Justify your choice.

a.) Given that $\displaystyle y = e^{-t}(\cos{t} + \sin{2t})$ solves a 2nd order, linear, homogeneous ordinary Dif EQ (ODE), the ODE in equation would be of the form $\displaystyle ay'' + by' + cy = 0$ where $\displaystyle a,b,c \in \mathbb{R}$ are constants.

b.) Given that $\displaystyle y_1$ and $\displaystyle y_2$ are fundamental sol'ns to a 2nd order, linear, homogenous ordinary dif. eq (ODE) with each solution defined on an open interval $\displaystyle I$. On all $\displaystyle I$, the Wronskian $\displaystyle W(y_1,y_2)$ is either strictly positive or it is strictly negative.

2. (a) False, general solution is $\displaystyle y = c_1e^{\lambda t} \cos(\mu t) + c_2e^{\lambda t} \sin(\mu t)$. Here $\displaystyle t \neq 2t$.

(b) $\displaystyle W(y_1, y_2)(t) = W(y_1,y_2)(t_0)e^{-\int_{t_0}^{t} p(x) \ dx} = ce^{-\int p(t) \ dt}$

3. Originally Posted by tukeywilliams
(a) False, general solution is $\displaystyle y = c_1e^{\lambda t} \cos(\mu t) + c_2e^{\lambda t} \sin(\mu t)$. Here $\displaystyle t \neq 2t$.

(b) $\displaystyle W(y_1, y_2)(t) = W(y_1,y_2)(t_0)e^{-\int_{t_0}^{t} p(x) \ dx} = ce^{-\int p(t) \ dt}$
So for b, the wronskian has to be strictly positive, and so it's false? Since e^(anything) is always 0 or positive

Thanks for the help.

4. $\displaystyle c$ could be $\displaystyle 0$. So the wronskian can be $\displaystyle 0$.

5. Originally Posted by tukeywilliams
$\displaystyle c$ could be $\displaystyle 0$. So the wronskian can be $\displaystyle 0$.
Actually, isn't this true then? Because c could be negative, and hence the wronskian would be too

6. Originally Posted by caeder012
On all $\displaystyle I$, the Wronskian $\displaystyle W(y_1,y_2)$ is either strictly positive or it is strictly negative.
.

You can just add: On all $\displaystyle I$, the Wronskian $\displaystyle W(y_1,y_2)$ is either strictly positive, $\displaystyle 0$, or strictly negative. So the above statement is false because it does not have $\displaystyle 0$.

7. Got it. One more question, for the first one, you say $\displaystyle t \neq 2t$. What do you mean by this?

8. We need a $\displaystyle \cos \mu t$ and a $\displaystyle \sin \mu t$.

Instead we have $\displaystyle \cos t$ and $\displaystyle \sin 2t$. We don't have a $\displaystyle \cos t, \ \sin t$ or $\displaystyle \cos 2t, \ \sin 2t$.