 # Section 2.1 – Linear Equations; Method of Integrating Factors

Directions. The following are review problems for the section. It is recommended you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it.
Note: In problems 1 and 2, some of the parts use techniques from Section 2.2.

1. Find the general solution of the given differential equation.
1. $$y'+2ty=2te^{-t^2}$$
2. $$2\sqrt{x}\,y'=\sqrt{1-y^2}$$
3. $$ty'+y=3t\cos t,\qquad t>0$$​

1. $$y=(t^2+C)e^{-t^2}$$
2. $$y=\sin(\sqrt{x}+C),\quad$$ equilibrium solutions at $$y=\pm1$$
3. $$y=\dfrac{3(t\sin(t)+\cos(t))+C}{t}$$

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2. Find the solution to the initial value problem and the interval of validity in each case.
1. $$2\sqrt{x}\dfrac{dy}{dx}=\cos^2 y,\qquad$$ $$y(4)=\dfrac{\pi}{4}$$
2. $$\dfrac{dy}{dt}+\dfrac{2y}{t}=\dfrac{\cos t}{t^2}\qquad$$ $$y(1)=\dfrac{1}{2},\qquad$$ $$t>0$$​

1. $$y=\arctan(\sqrt{x}-1), \quad$$$$I.V.=(0,\infty)$$
2. $$y=\dfrac{\sin(t)+\dfrac{1}{2}-\sin(1)}{t^2}, \quad$$ $$I.V. =(0,\infty)$$

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3. 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. ​

$$y_0=\dfrac{11}{4}-\dfrac{1}{4}e^{10}$$

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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