Second order equations with constant coefficientsAQA A-Level Further Maths: Revision notes

To solve $y''+ay'+by=0$ with constant $a,b$, try $y=e^{mx}$. Then $y'=me^{mx}$ and $y''=m^2e^{mx}$, and since $e^{mx}\neq0$ we need $m^2+am+b=0,$ the auxiliary equation. Its roots decide the form of the general solution, which has two arbitrary constants because the equation is second order. Solve the quadratic by factorising, completing the square or the formula. Example: $y''-5y'+6y=0$ gives $m^2-5m+6=0$, so $m=2$ or $3$.

If the discriminant $a^2-4b>0$ there are distinct real roots $m_1,m_2$ and $y=Ae^{m_1x}+Be^{m_2x}.$ For $y''-5y'+6y=0$: $y=Ae^{2x}+Be^{3x}$. Both exponents are real, so the solution grows or decays without oscillating. If both roots are negative, $y\to0$ as $x\to\infty$.

If the discriminant is $0$ there is one repeated root $m$ and $y=(A+Bx)e^{mx}.$ For $y''+4y'+4y=0$: $(m+2)^2=0$, so $y=(A+Bx)e^{-2x}$. The factor $x$ is needed to supply a second independent solution. To find $A$ and $B$ differentiate with the product rule: $y'=\left(B+mA+mBx\right)e^{mx}$.

If the discriminant is negative the roots are a conjugate pair $m=\alpha\pm i\beta$ and $y=e^{\alpha x}\left(A\cos\beta x+B\sin\beta x\right).$ For $y''+2y'+5y=0$: $m=-1\pm2i$, so $y=e^{-x}(A\cos2x+B\sin2x)$. The real part $\alpha$ gives growth ($\alpha>0$) or decay ($\alpha<0$) and $\beta$ the angular frequency of the oscillation. If $\alpha=0$ the oscillation has constant amplitude, as in $y''+9y=0$ with $y=A\cos3x+B\sin3x$.

Match the discriminant $a^2-4b$ to the case:

• $>0$: $Ae^{m_1x}+Be^{m_2x}$
• $=0$: $(A+Bx)e^{mx}$
• $<0$: $e^{\alpha x}(A\cos\beta x+B\sin\beta x)$ Two constants need two conditions, usually $y$ and $\frac{dy}{dx}$ at one value of $x$. Differentiate the general solution, substitute both conditions and solve the simultaneous equations. Example: $y=Ae^{2x}+Be^{3x}$ with $y=1$ and $y'=0$ at $x=0$ gives $A+B=1$ and $2A+3B=0$, so $y=3e^{2x}-2e^{3x}$.