Second-order linear differential equationsEdexcel A-Level Further Maths: Revision notes

A second-order linear equation with constant coefficients has the form $y''+ay'+by=f(x)$. When $f(x)=0$ it is homogeneous. Try $y=e^{mx}$: then $y'=me^{mx}$ and $y''=m^2e^{mx}$, giving the auxiliary equation $m^2+am+b=0.$ Its roots decide the form of the solution. The general solution of a second-order equation has two arbitrary constants. Example: $y''+2y'-8y=0$ gives $m^2+2m-8=0$, so $m=2$ or $m=-4$ and $y=Ae^{2x}+Be^{-4x}$.

The sign of the discriminant $a^2-4b$ decides the form of the complementary function:

• $a^2-4b>0$: two distinct real roots $\alpha,\beta$: $y=Ae^{\alpha x}+Be^{\beta x}$.
• $a^2-4b=0$: one repeated root $\alpha$: $y=(A+Bx)e^{\alpha x}$.
• $a^2-4b<0$: complex roots $p\pm q\mathrm{i}$: $y=e^{px}(A\cos qx+B\sin qx)$. Example: $y''+4y'+13y=0$ has $m=-2\pm3\mathrm{i}$, so $y=e^{-2x}(A\cos3x+B\sin3x)$. If $p=0$ the solution is pure oscillation, $A\cos qx+B\sin qx$, for example $y''+16y=0$ has $y=A\cos4x+B\sin4x$.

For $y''+ay'+by=f(x)$: $\text{general solution}=\text{complementary function (CF)}+\text{particular integral (PI)}.$ Find the CF as above, then find a PI by trying a function of the same form as $f(x)$:

• $f(x)=ke^{px}$: try $y=\lambda e^{px}$.
• $f(x)=A+Bx$ or $p+qx+cx^2$: try a polynomial of the same degree, $y=ax+b$ or $y=ax^2+bx+c$.
• $f(x)=m\cos\omega x+n\sin\omega x$: try $y=a\cos\omega x+b\sin\omega x$ (both terms, even if $f$ has only one). Substitute into the equation and compare coefficients.

Polynomial: $y''-3y'+2y=4x+2$. CF: $m^2-3m+2=0$, $m=1,2$, so $Ae^x+Be^{2x}$. Try $y=ax+b$: $-3a+2ax+2b=4x+2$, so $a=2$, $b=4$. General solution $y=Ae^x+Be^{2x}+2x+4$. Trigonometric: $y''+3y'+2y=10\sin x$. Try $y=a\cos x+b\sin x$: $y'=-a\sin x+b\cos x$, $y''=-a\cos x-b\sin x$. Then $(a+3b)\cos x+(b-3a)\sin x=10\sin x$, so $a+3b=0$ and $b-3a=10$, giving $a=-3$, $b=1$. PI: $-3\cos x+\sin x$. Initial conditions: apply them to the general solution (CF + PI), not to the CF alone. Use $y(0)$ for one equation and $y'(0)$ for the other, remembering the product rule when differentiating terms such as $xe^{3x}$.

If the trial form for the PI already appears in the CF, substituting gives 0 and fails. Multiply the trial function by $x$. Example: $y''-5y'+6y=4e^{3x}$. The CF is $Ae^{2x}+Be^{3x}$, which contains $e^{3x}$, so $y=ke^{3x}$ fails. Try $y=kxe^{3x}$: $y'=k(1+3x)e^{3x}$, $y''=k(6+9x)e^{3x}$, and substituting gives $ke^{3x}\left[(6+9x)-5(1+3x)+6x\right]=ke^{3x}$. Hence $k=4$ and the general solution is $y=Ae^{2x}+Be^{3x}+4xe^{3x}$. With a repeated root and $f(x)$ of that exponential form, multiply by $x^2$ instead.