CP 9.4, 9.6 - Second-order homogeneous equations

The auxiliary equation and distinct real roots

The equation $y''+4y=0$ has the real solution $y=\cos2x$. Yet trying an exponential $y=\mathrm{e}^{mx}$ asks for $m^2=-4$. How can complex roots lead to a real sine or cosine curve? The same exponential trial solves the real-root cases first; Euler's formula will then complete the picture.

A second-order linear homogeneous equation with constant coefficients has the form $y''+ay'+by=0$, with real constants $a,b$. “Homogeneous” refers to the zero right side. The equation $y''+ay'+by=f(x)$ with non-zero $f$ needs a further step, taught next.

Trying $y=\mathrm{e}^{mx}$ gives

The exponential never vanishes, so the bracket must be zero. Linearity means that a constant multiple or sum of solutions is also a solution. Two independent solutions provide the two constants in the general solution, also called the complementary function, $y_{\mathrm{CF}}$.

The independent variable belongs in both exponentials. An expression such as $A\mathrm{e}^{m_1}+B\mathrm{e}^{m_2}$ would just be a constant.

The second example changes only the coefficient of $y$. One auxiliary root now becomes negative, producing a decreasing exponential alongside an increasing one.

An auxiliary root of zero is allowed: $\mathrm{e}^{0x}=1$, so its term is a constant. The discriminant identifies the root type, while the roots' signs decide growth or decay.

Try another quadratic with opposite-sign roots.

A repeated root needs an extra factor of x

When the quadratic has a repeated root, writing the same exponential twice supplies only one independent term.

If $m_1=m_2=m$, then $A\mathrm{e}^{mx}+B\mathrm{e}^{mx}=(A+B)\mathrm{e}^{mx}$: the two constants have merged. Instead, set $y=\mathrm{e}^{mx}u(x)$. A repeated root means the original equation is $y''-2my'+m^2y=0$. Substitution cancels the terms in $u$ and $u'$, leaving $\mathrm{e}^{mx}u''=0$. Hence $u''=0$, so $u=A+Bx$.

There is a neat reason the factor $x$ appears when roots merge. For two nearby exponentials,

The apparently new solution is the limiting difference of the two old ones. This limit explains the extra factor; you can use the repeated-root form directly in a calculation.

The second example changes only the sign of the first-derivative coefficient. The root changes from $2$ to $-2$, while the repeated-root structure remains.

Even the term $x\mathrm{e}^{-2x}$ tends to zero as $x\to\infty$: the exponential decay dominates the linear factor. A polynomial multiplier changes the shape without automatically changing the long-term limit.

Change the repeated root in the final check for this case.

Complex roots give real trigonometric solutions

A negative discriminant gives conjugate roots; Euler's formula turns their exponentials into real sine and cosine terms.

Suppose $m=\alpha\pm\mathrm{i}\beta$, with $\beta>0$. Then

Their sum produces a real cosine term; their difference divided by $\mathrm{i}$ produces a real sine term. Use arbitrary real coefficients for these two real basis functions. Thus the real-valued solution has two arbitrary real constants multiplying cosine and sine.