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MATH
[1251]

Methods from Second-order constant-coefficient equations

All methods from Second-order constant-coefficient equations.

Read the characteristic roots

Solve the characteristic equation and choose the real homogeneous modes according to the roots:

  • If r1≠r2r_1\neq r_2 are distinct real roots, use yh=Aer1x+Ber2x.y_h=Ae^{r_1x}+Be^{r_2x}.
  • If rr is a repeated real root, use yh=(A+Bx)erx.y_h=(A+Bx)e^{rx}.
  • If the roots are r=α±iβr=\alpha\pm i\beta with β≠0\beta\neq0, use yh=eαx(Acos⁡(βx)+Bsin⁡(βx)).y_h=e^{\alpha x} \bigl(A\cos(\beta x)+B\sin(\beta x)\bigr).

For a repeated root, erxe^{rx} and xerxxe^{rx} are independent; the factor xx is required. The extra mode comes from writing y=erxv(x)y=e^{rx}v(x), where rr is the repeated root. Since b=−2arb=-2ar and c=ar2c=ar^2, substitution reduces the ODE to aerxv′′=0ae^{rx}v''=0, so v=A+Bxv=A+Bx.

For a complex pair, α\alpha controls exponential growth or decay and β\beta controls oscillation, as with the modulus and argument in polar form. For real coefficients, a non-real root always occurs with its complex conjugate. Since e(α+iβ)x=eαx(cos⁡βx+isin⁡βx)e^{(\alpha+i\beta)x}=e^{\alpha x}(\cos\beta x+i\sin\beta x), its real and imaginary parts give the two real modes in the last row and the final solution needs no complex constants.

Undetermined coefficients

For suitable forcing functions:

  • a polynomial of degree nn suggests a general polynomial of degree nn;
  • eβxe^{\beta x} suggests CeβxCe^{\beta x};
  • pn(x)eβxp_n(x)e^{\beta x} suggests qn(x)eβxq_n(x)e^{\beta x}, where qnq_n has degree nn;
  • sin⁡(βx)\sin(\beta x) or cos⁡(βx)\cos(\beta x) suggests C1cos⁡(βx)+C2sin⁡(βx);C_1\cos(\beta x)+C_2\sin(\beta x);
  • eαxsin⁡(βx)e^{\alpha x}\sin(\beta x) or eαxcos⁡(βx)e^{\alpha x}\cos(\beta x) suggests eαx(C1cos⁡(βx)+C2sin⁡(βx)).e^{\alpha x} \bigl(C_1\cos(\beta x)+C_2\sin(\beta x)\bigr).

Substitute the trial into the equation and match coefficients. If the trial overlaps with a homogeneous mode whose characteristic root has multiplicity ss, multiply the whole trial by xsx^s (so a simple overlap requires one factor of xx, while a repeated root requires x2x^2).

The trial families work because differentiating a polynomial, exponential or sinusoidal expression keeps it within the same finite-dimensional family, so substitution reduces the problem to algebraic coefficient matching. The correction accounts for resonance: a trial that already lies in the homogeneous solution space cannot produce the forcing after applying LL. Oscillation by itself is not resonance; resonance requires a forcing term whose trial shape overlaps a natural homogeneous mode.

Recover an IVP from an integral equation

Differentiate an integral equation with a variable limit, record the initial values obtained at the endpoint and solve the resulting differential equation subject to those values. Finally, check the original integral equation because differentiation can enlarge the solution set.

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