Author Topic: TUT0303 Quiz4  (Read 629 times)

Tiantian Yu

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TUT0303 Quiz4
« on: October 18, 2019, 02:00:00 PM »
Find the general solution of the given differential equation: y''+4y'+6.25y=0

solution: The auxiliary equation is \begin{equation*}r^2+4r+6.25=0\end{equation*}
By the quadratic formula, the roots are \begin{equation*}r=\frac{-4\pm\sqrt{4^2 - 4(6.25)}}{2}=-2\pm\frac{3}{2}i.\end{equation*}
If the roots of the auxiliary equation are the complex numbers 􏲬\begin{equation*}r=\alpha\pm i\beta\end{equation*} then the general solution of ay''+by'+cy=0 is\begin{equation*} y=e^{\alpha t} (C_1 cos\beta t +C_2sin\beta t)\end{equation*}
since􏲬\begin{equation*}\alpha=-2,\beta=\frac{3}{2}\end{equation*}
the general solution of the differential equation is \begin{equation*} y=e^{-2 t} (C_1 cos\frac{3}{2} t +C_2sin\frac{3}{2} t)\end{equation*}