Constant coefficient second order equationsEdexcel International A Level Further Maths: Revision notes

The linear second order equation with constant coefficients is $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=f(x),$ where $a,b,c$ are real constants ($a\neq0$). The general solution is $y=\text{complementary function}+\text{particular integral}.$

• The complementary function (CF) is the general solution of the homogeneous equation $a y''+by'+cy=0$ and contains two arbitrary constants.
• The particular integral (PI) is any one solution of the full equation and contains no arbitrary constants. The two constants are found from two conditions, usually the value of $y$ and of $\frac{dy}{dx}$ at one point.

Try $y=e^{mx}$ in the homogeneous equation to get the auxiliary equation $am^2+bm+c=0$. The roots decide the CF:

• Two distinct real roots $m_1,m_2$: $y=Ae^{m_1x}+Be^{m_2x}$.
• One repeated root $m$ (discriminant $0$): $y=(A+Bx)e^{mx}$.
• Complex roots $\alpha\pm i\beta$: $y=e^{\alpha x}(A\cos\beta x+B\sin\beta x)$. Examples: $y''-7y'+12y=0$ has $m=3,4$, so $y=Ae^{3x}+Be^{4x}$. $y''-4y'+4y=0$ has $m=2$ repeated, so $y=(A+Bx)e^{2x}$. $y''+2y'+10y=0$ has $m=-1\pm3i$, so $y=e^{-x}(A\cos3x+B\sin3x)$. Complex roots give oscillations with an envelope $e^{\alpha x}$: they decay when $\alpha<0$ and grow when $\alpha>0$.

Choose a trial form matching $f(x)$, substitute it into the full equation, and compare coefficients.

• $f(x)=ke^{px}$: try $\lambda e^{px}$.
• $f(x)=A+Bx$: try $\lambda+\mu x$.
• $f(x)=p+qx+cx^2$: try $\lambda+\mu x+\nu x^2$.
• $f(x)=m\cos\omega x+n\sin\omega x$: try $\lambda\cos\omega x+\mu\sin\omega x$ (both terms, even if $f$ has only one). Example: $y''-3y'+2y=e^{3x}$. Try $\lambda e^{3x}$: $9\lambda-9\lambda+2\lambda=1$, so $\lambda=\frac12$ and the PI is $\frac12e^{3x}$. Example: $y''+y'-2y=x$. Try $\lambda+\mu x$: $\mu-2\lambda-2\mu x=x$, so $\mu=-\frac12$, $\lambda=-\frac14$.

If the trial form already appears in the CF, it satisfies the homogeneous equation and gives $0$, so it cannot match $f(x)$. Multiply the trial form by $x$ (and by $x^2$ if the CF contains a repeated root of the same type). Example: $y''+9y=\cos3x$. The CF is $A\cos3x+B\sin3x$, which contains $\cos3x$. Try $y=x(p\cos3x+q\sin3x)$. Differentiating twice gives $y''+9y=-6p\sin3x+6q\cos3x$, so $p=0$ and $q=\frac16$. The PI is $\frac x6\sin3x$. The same applies to $y''+4y=\sin2x$ (CF contains $\sin2x$). The growing factor $x$ means the oscillations increase in amplitude without bound, called resonance.

To find the constants, write the general solution, differentiate it (product rule where there is a factor $x$), and substitute the two conditions. Example: $y''-3y'+2y=e^{3x}$ with $y=1$ and $y'=0$ at $x=0$. The general solution is $y=Ae^x+Be^{2x}+\frac12e^{3x}$. Then $A+B+\frac12=1$ and $A+2B+\frac32=0$, so $B=-2$ and $A=\frac52$, giving $y=\frac52e^x-2e^{2x}+\frac12e^{3x}$. Check by substituting back into the equation and the conditions. In context, say what happens as $x$ becomes large: decaying terms vanish and the PI describes the long-term behaviour.