# Thread: Dif EQ T/F

1. ## Dif EQ T/F

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

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

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

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

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

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

(b) $W(y_1, y_2)(t) = W(y_1,y_2)(t_0)e^{-\int_{t_0}^{t} p(x) \ dx} = ce^{-\int p(t) \ dt}$
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. $c$ could be $0$. So the wronskian can be $0$.

5. Originally Posted by tukeywilliams
$c$ could be $0$. So the wronskian can be $0$.
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 $I$, the Wronskian $W(y_1,y_2)$ is either strictly positive or it is strictly negative.
.

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

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

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

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