Revise Differential equations for Cambridge 9231 (2.6)

A first-order linear differential equation can be written in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$. To solve these, we use a clever technique involving an 'integrating factor'. This factor is specifically constructed to turn the left-hand side of the equation into the result of the product rule for differentiation, which can then be easily integrated.

Integrating Factor, $I(x) = e^{\int P(x) dx}$

Once you've calculated the integrating factor, $I(x)$, you multiply every term in the standard form equation by it. The equation then simplifies to $\frac{d}{dx}(I(x)y) = I(x)Q(x)$. From here, you can integrate both sides with respect to $x$ to find an expression for $y$.

These equations have the form $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0$, where $a, b,$ and $c$ are constants. The key to solving them is the auxiliary equation, $am^2 + bm + c = 0$. The nature of the roots of this quadratic equation dictates the form of the general solution, which is called the Complementary Function (CF).

When the right-hand side is not zero, i.e., $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)$, the equation is non-homogeneous. The general solution is the sum of two parts: the Complementary Function (CF) from the homogeneous version, and a Particular Integral (PI), which is a specific solution that depends on the form of $f(x)$.

To find the PI, we try a function of a similar form to $f(x)$. For example, if $f(x)$ is a quadratic, we try a generic quadratic for the PI. If $f(x)$ is an exponential, we try a generic exponential. We then substitute this trial function into the original differential equation to determine the specific coefficients.