Author Topic: TUT0102 Quiz3  (Read 352 times)

lilywq

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TUT0102 Quiz3
« on: October 11, 2019, 03:02:25 PM »
\begin{align*}
 y^"+8y^{'}-9y=0 && {y(1)=1,y'(1)=0}\\
 r^2+8r-9&=0\\
 (r+9)(r-1)&=0\\
 r=-9 \text{ or } r=1\\
 y&=c_{1}e^{-9t}+c_{2}e^t\\
 y(1)&=c_1e^{-9(1)}+c_{2}e^{1}\\
 1&=c_{1}e^{-9t}+c_{2}e^t\\
 \text{Differentiate y with respect to t, we get}\\
 y^{'}&= -9c_{1}e^{-9t}+c_{2}e^{t}\\
 y^{'}(1)&= -9c_{1}e^{-9(1)}+c_{2}e^{1}\\
 0 &= -9c_{1}e^{-9}+c_{2}e\\
 9c_{1}e^{-9}&=c_{2}e\\
 c_{1}&=\frac{c_{2}e^{10}}{9}\\
  \text{Substitute } c_{1}=\frac{c_{2}e^{10}}{9} \text{ in } 1=c_{1}e^{-9}+c_{2}e\\
  1&=(\frac{c_{2}e^{10}}{9})e^{-9}+c_{2}e\\
  1&=\frac{10}{9}c_{2}e\\
  c_{2}&=\frac{9}{10e}\\
  c_{1}&=\frac{1}{10}e^9\\
  \text{Therefore, the general solution of the initial value problem(1) is}\\
  y=\frac{1}{10}e^{9(1-t)}+\frac{9}{10}e^{t-1}
\end{align*}