quiz 2 (tut 0301)
$$ y’’ + 4y = 0,\ y_{1}(t) = \cos(2t),\ y_{2}=\sin(2t)$$
The given solution of differential equation is: $y_{1}(t) = \cos(2t)$
So, $y_{1}’(t) = -2 \sin(2t)$
$y_{1}’’(t) = -4 \cos(2t)$
Then, plug into the given differential equation: $y’’ + 4y = 0$,
i.e. $-4 \cos(2t)\ + 4(\cos(2t)) = 0$ => 0=0
So, $y_{1}(t) = \cos(2t)$ is a solution of this equation
Similarly, we have $y_{2}(t)=\sin(2t)$
So, $y_{2}’(t) = 2 \cos(2t)$
$y_{2}’’(t) = -4 \sin(2t)$
Then, plug into the given differential equation: $y’’ + 4y = 0$,
i.e. $-4 \sin(2t)\ + 4(\sin(2t)) = 0$ => 0=0
So, $y_{2}(t)=\sin(2t)$ is also a solution of this equation
To check whether $y_{1}$ and $y_{2}$ constitute a fundamental set of solutions, we will find the Wronskian $W(y_{1}(t), y_{2})(t)\ = \begin{vmatrix}
y_{1}(t) & y_{2}(t)\\
y_{1}’(t) & y_{2}’(t)\\
\end{vmatrix} $
= $ \begin{vmatrix}
\cos(2t) & \sin(2t)\\
-2\sin(2t) & 2\cos(2t)\\
\end{vmatrix} $
= $2\cos^2(2t)+2\sin^2(2t)$
= $2(\cos^2(2t)+\sin^2(2t))$
= $2 (1)$
= $2$
Since, $W(y_{1}(t), y_{2})(t)\neq0$, then we say that $y_{1}$ and $y_{2}$ constitute a fundamental set of solutions.
