Video Errata: At the 10:30 mark, the presenter wrote \(\dfrac{11}{4}-\dfrac{t_1}{4}+(y_0-\dfrac{11}{4})e^{-2t_1}=0\) but we should have \(\dfrac{11}{4}-\dfrac{t_1}{2}+(y_0-\dfrac{11}{4})e^{-2t_1}=0\). Continuing from there, we still get that \(t_1=5\), but now we have the equation \(\dfrac{11}{4}-\dfrac{5}{2}+(y_0-\dfrac{11}{4})e^{-10}=0\). Solving for \(y_0\) then gives us that \(y_0=\dfrac{11}{4}-\dfrac{1}{4}e^{10}\).
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MATH 308 WIR22A V11