First-Order Linear Differential Equations

Professor Gilbert Strang gives an excellent explanation in this MIT OpenCourseWare lesson. If you are interested in differential equations, I highly recommend it.

Linearity

\begin{align} y^{\prime}(x) + P(x)y(x) = Q(x) \end{align}

We do not yet know how to solve a differential equation in this form, but suppose it could be written as:

\begin{align} \mu y^{\prime} + \mu ^{\prime}y = Q(x)\nonumber \end{align}

The product rule then gives:

\begin{align} \mu y^{\prime} + \mu ^{\prime}y = Q = \frac{d}{dx}(\mu y)\nonumber \end{align}

Finally, integrate both sides and rearrange:

\begin{align} \mu y &= \int Q\mathbf{d}x\nonumber\\ y &= \frac{1}{\mu}\int Q\mathbf{d}x\nonumber \end{align}

Therefore, construct $\mu$ so that $\mu^{\prime} = \mu P$ , and solve the homogeneous equation for $\mu$ :

\begin{align} \frac{\mu^{\prime}}{\mu} &= P\nonumber\\ ln(\mu) &= \int P\mathbf{d}x + a\nonumber\\ \mu &= Ce^{\int P\mathbf{d}x} \end{align}

Multiply both sides of $(1)$ by $\mu$ , then integrate both sides:

\begin{align} \mu (y^{\prime}+Py) &= \mu Q\nonumber\\ \mu y^{\prime} + \mu Py &= \mu Q \nonumber\\ \mu y^{\prime} + \mu^{\prime}y &= \mu Q\nonumber\\ \mu y &= \int \mu Q \mathbf{d}x + C\nonumber\\ y &= \frac{1}{\mu} (\int \mu Q \mathbf{d}x + C) \end{align}

If the original equation is homogeneous, that is, $Q(x) = 0$ , then

\begin{align} y_h &= \frac{1}{\mu}C\nonumber \end{align}

If the original equation is nonhomogeneous, we can choose $C = 0$ in particular, giving

\begin{align} y_p &= \frac{1}{\mu}\int \mu Q \mathbf{d}x\nonumber \end{align}

Looking at the two terms on either side of the plus sign in $(3)$ , we can see that the general solution to a first-order linear differential equation consists of the homogeneous solution and a particular solution: $y = y_p + y_h$ .

First-Order Linear Differential Equations

Linearity

Separable Equations

Homogeneous Equations

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