Series solutions and reducible differential equationsEdexcel A-Level Further Maths: Revision notes

To find a series solution of a differential equation when an exact solution is hard, assume $y=y(0)+xy'(0)+\frac{x^2}{2!}y''(0)+\frac{x^3}{3!}y'''(0)+\dots$ (a Taylor series about $0$) and find the derivatives at $x=0$ from the equation itself. Method: (1) write $y''$ (or $y'$) in terms of $x,y,y'$; (2) differentiate repeatedly, using the product rule and the chain rule; (3) substitute $x=0$ and the initial conditions; (4) assemble the series. Each differentiation gives the next derivative.

Solve $y''+xy'+y=0$ with $y(0)=1$, $y'(0)=0$ as a series to $x^4$. Rearranging, $y''=-xy'-y$, so $y''(0)=-1$. Differentiate: $y'''=-y'-xy''-y'=-2y'-xy''$, so $y'''(0)=0$. Differentiate again: $y^{(4)}=-2y''-y''-xy'''=-3y''-xy'''$, so $y^{(4)}(0)=3$. $y=1+\frac{x^2}{2!}(-1)+\frac{x^4}{4!}(3)=1-\frac{x^2}{2}+\frac{x^4}{8}.$ The odd terms vanish because $y'(0)=0$ and $y'''(0)=0$.

Series for $\frac{\mathrm{d}y}{\mathrm{d}x}=x-y^2$ with $y(0)=1$. At $x=0$: $y'=-1$. Differentiating gives $y''=1-2yy'$, so $y''(0)=1+2=3$. Then $y'''=-2(y')^2-2yy''$, so $y'''(0)=-2-6=-8$. The series is $y=1-x+\frac32x^2-\frac43x^3+\dots$ Nonlinear equations with no closed-form solution are typical for this method.

Some equations can be changed by a given substitution into one of the standard types from Core Pure: a first-order linear equation, solved with an integrating factor, or a second-order linear equation with constant coefficients, solved with a complementary function and a particular integral. After solving, substitute back to return to the original variable. The question tells you which substitution to use; the work is in transforming the derivatives correctly using the chain rule.

Solve $x\frac{\mathrm{d}y}{\mathrm{d}x}+y=xy^2$ using $z=\frac1y$. Then $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{z^2}\frac{\mathrm{d}z}{\mathrm{d}x}$, giving $-\frac{x}{z^2}z'+\frac1z=\frac{x}{z^2}$ and so $z'-\frac zx=-1$. The integrating factor is $\mathrm{e}^{-\int\frac1x\,\mathrm{d}x}=\frac1x$, so $\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac zx\right)=-\frac1x$ and $\frac zx=-\ln x+c$. Therefore $z=x(c-\ln x)$ and $y=\frac{1}{x(c-\ln x)}$. With $y(1)=1$, $c=1$.

For $x^2y''-2xy'+2y=4x^3$ with $x=\mathrm{e}^t$ (so $t=\ln x$): $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac1x\frac{\mathrm{d}y}{\mathrm{d}t}$ and $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{1}{x^2}\left(\frac{\mathrm{d}^2y}{\mathrm{d}t^2}-\frac{\mathrm{d}y}{\mathrm{d}t}\right)$. The equation becomes $\ddot y-3\dot y+2y=4\mathrm{e}^{3t}$ (dots for $\frac{\mathrm{d}}{\mathrm{d}t}$). Auxiliary equation $m^2-3m+2=0$, so $m=1,2$ and the complementary function is $A\mathrm{e}^t+B\mathrm{e}^{2t}$. For the particular integral try $k\mathrm{e}^{3t}$: $9k-9k+2k=4$, so $k=2$. Hence $y=A\mathrm{e}^t+B\mathrm{e}^{2t}+2\mathrm{e}^{3t}=Ax+Bx^2+2x^3$.