Section 2.1 – Linear Equations; Method of Integrating Factors
Exercises
Directions: You should try to solve each problem first, and then click "Reveal Answer" to check your answer. You can click "Watch Video" if you need help with a problem.
1. Find the general solution of the differential equation: \(y'+2ty=2te^{-t^2}\)
\(y=(t^2+C)e^{-t^2}\)
2. Find the general solution of the differential equation: \(ty'+y=3t\cos t,\qquad t>0\)
\(y=\dfrac{3(t\sin(t)+\cos(t))+C}{t}\)
3. Find the solution to the initial value problem and the interval of validity:
\(\dfrac{dy}{dt}+\dfrac{2y}{t}=\dfrac{\cos t}{t^2}\qquad\) \(y(1)=\dfrac{1}{2},\qquad\) \(t>0\)
\(y=\dfrac{\sin(t)+\dfrac{1}{2}-\sin(1)}{t^2}, \quad\) \(I.V. =(0,\infty)\)
4. Consider the initial value problem \[y'+2y=5-t,\qquad y(0)=y_0.\] Find the value \(y_0\) for which the solution touches, but does not cross the \(t\)-axis.
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 \(y_0=\dfrac{11}{4}-\dfrac{1}{4}e^{10}\).
\(y_0=\dfrac{11}{4}-\dfrac{1}{4}e^{10}\)