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### First Order Linear Equations

#### Additional Examples

Solve the following initial value problem
$$
\begin{align}
\frac{dy}{dx} - 2 y &= 2\cos(3 x) \\
y(0) &= 4
\end{align}
$$
This is a linear equation. First we find the general solution following the paradigm.
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©2010, 2014 Andrew G. Bennett

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- Find the integrating factor $$ \mu(x) = \exp(\int -2 dx) = \exp(-2x) $$
- Multiply through by the integrating factor $$ \exp(-2x) \frac{dy}{dx} - 2\exp(-2x)y = 2\cos(3 x)\exp(-2x) $$
- Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)y).$ $$ \frac{d}{dx}(\exp(-2x)y) =2\cos(3 x)\exp(-2x) $$
- Integrate both sides. In this case you will need to integrate by parts twice and then solve for the unknown integral to evaluate the integral on the right (or you can use a table of integrals - see formula 97).$$ \exp(-2x)y = ((6/13)\sin(3x) - (4/13)\cos(3x))\exp(-2x) + C $$
- Divide through by $\mu(x)$ to solve for $ y.$ $$y = (6/13)\sin(3x) - (4/13)\cos(3x) + C\exp(2x) $$

If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett