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### First Order Linear Equations

Solve the following initial value problem \begin{align} \frac{dy}{dx} + 3 y &= -6 x + 2 \\ y(0) &= -3 \end{align} This is a linear equation. First we find the general solution following the paradigm.
1. Find the integrating factor $$\mu(x) = \exp(\int 3 dx) = \exp(3x)$$
2. Multiply through by the integrating factor $$\exp(3x) \frac{dy}{dx} + 3\exp(3x)y = (-6 x + 2)\exp(3x)$$
3. Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)y).$ $$\frac{d}{dx}(\exp(3x)y) =(-6 x + 2)\exp(3x)$$
4. Integrate both sides. In this case you will need to integrate by parts to evaluate the integral on the right.$$\exp(3x)y = (-2x + 4/3)\exp(3x) + C$$
5. Divide through by $\mu(x)$ to solve for $y.$ $$y = -2x + 4/3+ C\exp(-3x)$$
Now we plug in the initial values $x = 0$ and $y = -3$ and solve for $C = -13/3$, to obtain the solution to the initial value problem $$y = -2x + 4/3 - (13/3)\exp(-3x)$$ You may reload this page to generate additional examples.