Second-order differential equations

What you’ll learn

• Construct complementary functions from auxiliary roots.
• Find particular integrals and apply two conditions.

Understand the idea

For a linear equation with constant coefficients, substitute $y=e^{mx}$ into its homogeneous part to obtain an auxiliary quadratic. Distinct real roots give two exponentials; a repeated root introduces an $x$ factor; complex conjugate roots give exponentially scaled sine and cosine terms.

A nonzero right-hand side requires a particular integral as well as the complementary function. Try a form matching the forcing term and determine its coefficients. If that trial duplicates a complementary-function term, multiply by a sufficient power of $x$. Two independent conditions determine the two remaining arbitrary constants.

Key formulae

$$ay''+by'+cy=f(x),\qquad am^2+bm+c=0$$

$$y_c=Ae^{m_1x}+Be^{m_2x}\quad(m_1\ne m_2)$$

$$y_c=(A+Bx)e^{mx}\quad\text{(repeated root)}$$

$$y_c=e^{\alpha x}(A\cos\beta x+B\sin\beta x)\quad(m=\alpha\pm i\beta)$$

Worked example

Solve $y''-3y'+2y=0$ with $y(0)=3$ and $y'(0)=4$.

1. The auxiliary equation is $(m-1)(m-2)=0$.
2. Thus $y=Ae^x+Be^{2x}$.
3. The conditions give $A+B=3$ and $A+2B=4$.
4. Subtract to get $B=1$, then $A=2$.

Answer: $y=2e^x+e^{2x}$.