Linear ODEs

Linear ODEs always take the form of

$$\frac{dy}{dx} + y\cdot P(x) = Q(x)$$

When $Q(x) = 0$ this becomes a separable ODE and can be solved as before such that

$$dy \cdot \frac{1}{y} = -P(x) dx$$

$$y = Ae^{\int -P(x)dx}\quad\textrm{where } A = e^c$$

When $P(x) = 0$ this becomes a simple separable ODE and can be solved by integration.

Recall the product rule $\frac{d}{dx}(u\cdot v) = u'v + uv'$
This is strikingly close to our Linear ODE, except we are missing a factor, the integration factor, $\mu(x)$. We want $\mu$ to be such that $\mu' = P(x)\mu$, allow us to do some algebraic manipulation.

$$\mu' = P(x)\mu$$

$$P(x) = \frac{\mu'}{\mu} = \frac{d}{dx}\ln(\mu)$$

$$\int P(x) dx = \ln(\mu)$$

$$\mu = e^{\int P(x) dx}$$

We can then multiply by our integrating factor, then apply the chain rule, however this is best shown with an example.

$$\cos(x)\frac{dy}{dx} + \sin(x)y = 1$$

$$\frac{dy}{dx} + \tan(x)y = \sec(x) \quad\textrm{Linear ODE}$$

$$\mu = e^{\int \tan(x) dx}$$

$$\int \tan(x) dx = \ln(sec(x))$$

$$\mu = \sec(x) \quad\textrm{Multiply ODE by $\mu$}$$

$$\frac{dy}{dx}\sec(x) + \sec(x)\tan(x)y = \sec^2(x)$$

Since $\frac{d}{dx}\sec(x) = \sec(x)\tan(x)$ we should now be able to see how the LHS of our equation can be withdrawn into its chain rule form.

$$\frac{d}{dx}\sec(x)\cdot y = \frac{dy}{dx}\sec(x) + \sec(x)\tan(x)y$$

$$\frac{d}{dx}\sec(x)\cdot y = \sec^2(x)$$

$$\sec(x) \cdot y = \int \sec^2(x) = \tan(x) + c$$

$$y = \sin(x) + c\cdot \cos(x)$$