Author Topic: TUT0303 Quiz3  (Read 254 times)

Tiantian Yu

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TUT0303 Quiz3
« on: October 11, 2019, 02:00:39 PM »
Verify that the functions y1 and y2 are solutions of the given differential equation. Do they constitute a fundamental set of solutions?
y'' + 4y = 0;  y1(t)=cos(2t), y2(t)=sin(2t)

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solution: firstly verifying that the functions y1 and y2 are solutions of the given differential equation
verify y1(t)=cos(2t):
Left-hand side=y'' + 4y=(cos(2t))''+4(cos(2t))
                                  =(-2sin(2t))'+4(cos(2t))
                                  =-4cos(2t)+4(cos(2t))
                                  =0=Right-hand side

verify y2(t)=sin(2t):
Left-hand side=y'' + 4y=((sin(2t))''+4(sin(2t))
                                  =((2cos(2t))'+4(sin(2t))
                                  =(-4sin(2t)+4(sin(2t))
                                  =0=Right-hand side
Thus the functions y1 and y2 are solutions of the given differential equation.

Now we are checking if the given functions constitute a fundamental set of solutions
We calculate the Wronskian of y1 and y2:
\begin{equation*}
W=\begin{bmatrix}
cos(2t)  &  sin(2t)\\
-2sin(2t)  &   2cos(2t)
\end{bmatrix}\end{equation*}
\begin{equation*}=cos(2t)(2cos(2t))-[sin(2t)(-2sin(2t)]\end{equation*}
\begin{equation*}=2cos^2(2t)-(-2sin^2(2t))\end{equation*}
\begin{equation*}=2(cos^2(2t)+sin^2(2t))\end{equation*}
\begin{equation*}=2\end{equation*}
Since the Wronskian of y1 and y2 is not zero, which means they are linearly independent, the functions y1 and y2 constitute a fundamental set of solutions
« Last Edit: October 11, 2019, 06:18:51 PM by Tiantian Yu »