Further differential equations (Further Pure 1)

Explanation

• When a differential equation cannot be solved in closed form, a Taylor series about the point where the initial conditions are given still produces a usable local solution. The method is mechanical:
• substitute the initial values into the equation to get the first unknown derivative, differentiate the whole equation with respect to $x$, substitute again, and repeat until you have every derivative you need.
• Assemble them with $y=y(a)+y'(a)(x-a)+\dfrac{y''(a)}{2!}(x-a)^2+\cdots$. For a first-order equation $y'=f(x,y)$ you differentiate $f$ using the chain rule, remembering that $y$ is a function of $x$, so $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(y^2\right)=2yy'$.
• For a second-order equation both $y(a)$ and $y'(a)$ are given and the equation supplies $y''(a)$ directly.
• Keep a running table of the values at $x=a$; that table, not the algebra, is what you assemble the series from.

Explanation

• The question always supplies the substitution; your job is to carry it through correctly and land on one of the standard forms from Core Pure section 9 — a first-order linear equation solved by an integrating factor, a separable equation, or a second-order linear equation with constant coefficients. Three patterns dominate.
• First, $z=y^{1-n}$ turns $y'+P(x)y=Q(x)y^{n}$ into a linear equation in $z$; the working starts by dividing through by $y^{n}$.
• Second, $y=vx$ turns a homogeneous equation into a separable one in $v$ and $x$, using $\dfrac{\mathrm{d}y}{\mathrm{d}x}=v+x\dfrac{\mathrm{d}v}{\mathrm{d}x}$.
• Third, $x=\mathrm{e}^{u}$ turns an Euler equation into constant coefficients, using $x\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{\mathrm{d}y}{\mathrm{d}u}$ and $x^2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=\dfrac{\mathrm{d}^2y}{\mathrm{d}u^2}-\dfrac{\mathrm{d}y}{\mathrm{d}u}$.
• Always convert the final answer back to the original variables, and apply any boundary condition only after that conversion.