20 Jun 2025
Homogeneous Differential Equations with Constant Coefficients
Characteristic roots and the complete real solutions for distinct, repeated, and complex roots.
Consider the homogeneous constant-coefficient equation
\[a_n\frac{d^ny}{dx^n}+a_{n-1}\frac{d^{n-1}y}{dx^{n-1}} +\cdots+a_1\frac{dy}{dx}+a_0y=0, \qquad a_n\ne0.\]The exponential $y=e^{mx}$ is useful because every differentiation only multiplies it by $m$:
\[\frac{d^k}{dx^k}e^{mx}=m^ke^{mx}.\]Substitution gives
\[e^{mx}\left(a_nm^n+a_{n-1}m^{n-1}+\cdots+a_1m+a_0\right)=0.\]Since $e^{mx}\ne0$, the allowed values of $m$ satisfy the characteristic equation
\[\boxed{a_nm^n+a_{n-1}m^{n-1}+\cdots+a_1m+a_0=0}.\]Distinct real roots
For
\[y''-5y'+6y=0,\]the characteristic polynomial is
\[m^2-5m+6=(m-2)(m-3).\]The two independent solutions are $e^{2x}$ and $e^{3x}$, so
\[\boxed{y=C_1e^{2x}+C_2e^{3x}}.\]Repeated roots
If a root $m_0$ occurs $s$ times, the $s$ independent solutions are
\[e^{m_0x},\ xe^{m_0x},\ldots,x^{s-1}e^{m_0x}.\]For example,
\[y''-4y'+4y=0\]has $(m-2)^2=0$, and therefore
\[\boxed{y=(C_1+C_2x)e^{2x}}.\]To see why the factor $x$ appears, start with the distinct-root quotient
\[\frac{e^{(m_0+\varepsilon)x}-e^{m_0x}}{\varepsilon}.\]Taking $\varepsilon\to0$ gives
\[\frac{\partial}{\partial m_0}e^{m_0x}=xe^{m_0x}.\]Complex-conjugate roots
With real coefficients, a complex root $m=\alpha+i\beta$ is accompanied by $\alpha-i\beta$. Euler’s formula gives
\[e^{(\alpha+i\beta)x}=e^{\alpha x} \bigl(\cos\beta x+i\sin\beta x\bigr).\]Taking real linear combinations of the conjugate solutions yields
\[\boxed{y=e^{\alpha x} \left(C_1\cos\beta x+C_2\sin\beta x\right)}.\]For $y’‘+4y=0$, $m=\pm2i$, so
\[\boxed{y=C_1\cos2x+C_2\sin2x}.\]Each result is verified by direct differentiation in the Unit II Maxima worksheet.
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