Second order equations reducible by substitutionEdexcel International A Level Further Maths: Revision notes

Some second-order equations do not have constant coefficients, so the usual method (auxiliary equation, complementary function, particular integral) cannot be used directly. A substitution replaces either the variable $x$ by a new variable $t$ (change of independent variable) or $y$ by a new function $z$ (change of dependent variable) so that the new equation has constant coefficients. In this specification the substitution is always given. The steps are always the same:

1. Express each derivative of $y$ in terms of the new variable, using the chain rule or the product rule.
2. Substitute into the equation and simplify to a constant-coefficient equation.
3. Solve it by the standard method.
4. Convert back to the original variables and apply any initial or boundary conditions.

For $x>0$ the substitution $x=\mathrm{e}^t$ (so $t=\ln x$) is the standard one. A dot denotes $\frac{d}{dt}$. By the chain rule, since $\frac{dt}{dx}=\frac1x=\mathrm{e}^{-t}$: $\frac{dy}{dx}=\mathrm{e}^{-t}\dot y,\qquad\text{so}\qquad x\frac{dy}{dx}=\dot y.$ For the second derivative, differentiate $\mathrm{e}^{-t}\dot y$ with respect to $x$, using the product rule and then the chain rule again: $\frac{d^2y}{dx^2}=\mathrm{e}^{-t}\frac{d}{dt}\left(\mathrm{e}^{-t}\dot y\right)=\mathrm{e}^{-2t}\left(\ddot y-\dot y\right),\qquad\text{so}\qquad x^2\frac{d^2y}{dx^2}=\ddot y-\dot y.$ These two results turn any equation of the form $ax^2y''+bxy'+cy=f(x)$ (an Euler equation) into $a\ddot y+(b-a)\dot y+cy=f(\mathrm{e}^t)$, which has constant coefficients.

Sometimes the substitution replaces $y$ by a product, for example $y=z\,\mathrm{e}^{-x^2}$ where $z$ is a new function of $x$. Differentiate with the product rule and substitute. Example: with $y=z\mathrm{e}^{-x^2}$, $y'=(z'-2xz)\mathrm{e}^{-x^2},\qquad y''=\left(z''-4xz'+(4x^2-2)z\right)\mathrm{e}^{-x^2}.$ In $y''+4xy'+(4x^2+2)y$, the $z'$ terms cancel ($-4xz'+4xz'$) and the $z$ terms give $(4x^2-2)-8x^2+(4x^2+2)=0$, leaving $\mathrm{e}^{-x^2}z''$. The equation becomes $z''=0$ (or $z''=g(x)$ if the right-hand side is non-zero), which integrates directly. Every term contains the factor $\mathrm{e}^{-x^2}$, which is never zero, so it can be cancelled from both sides.

Once the equation has constant coefficients, solve $a\ddot y+b\dot y+cy=f(t)$ as in FP2 5.1:

• the auxiliary equation $am^2+bm+c=0$ gives the complementary function (two real roots $m_1,m_2$: $A\mathrm{e}^{m_1t}+B\mathrm{e}^{m_2t}$; repeated root $m$: $(A+Bt)\mathrm{e}^{mt}$; complex roots $p\pm iq$: $\mathrm{e}^{pt}(A\cos qt+B\sin qt)$);
• a particular integral is chosen to match $f(t)$; if its form already appears in the complementary function, multiply by $t$ (or $t^2$ for a repeated root);
• the general solution is the complementary function plus the particular integral. Finally substitute back using $t=\ln x$ and $\mathrm{e}^t=x$. For example $y=(A+Bt)\mathrm{e}^{-2t}$ becomes $y=\frac{A+B\ln x}{x^2}$.

Conditions are usually given in terms of $x$, so convert them carefully.

• A value $y(x_0)$ becomes $y$ at $t_0=\ln x_0$ (at $x=1$, $t=0$).
• A condition on $\frac{dy}{dx}$ becomes a condition on $\dot y$ through $\dot y=x\frac{dy}{dx}$. At $x=1$, $\dot y=\frac{dy}{dx}$. Worked example: $x^2y''+5xy'+4y=0$ with $y(1)=1$, $y'(1)=0$. With $x=\mathrm{e}^t$ the equation is $\ddot y+4\dot y+4y=0$, so $y=(A+Bt)\mathrm{e}^{-2t}$. At $t=0$: $A=1$. Then $\dot y=(B-2A-2Bt)\mathrm{e}^{-2t}$, equal to $0$ at $t=0$, so $B=2$. Hence $y=\frac{1+2\ln x}{x^2}$. Check the answer by differentiating it, or by testing the original equation at one value of $x$.