Ordinary Differential Equations

Wronskian and Abel's formula

For two solutions y_1 and y_2 of y''+p(x)y'+q(x)y=0 on an interval, the Wronskian is W(y_1,y_2)=y_1y_2'-y_1'y_2. The two solutions are linearly independent exactly when W is nonzero at some point of the interval. For this kind of equation, W is either identically zero or never zero on the interval.

Abel's formula gives the Wronskian without solving the equation: W(x)=W(x_0)\exp\left(-\int_{x_0}^{x}p(t)\,dt\right). The exponent carries a minus sign. A term p(x)=-\frac{3}{x} therefore produces growth like x^{3}, not x^{-3}. Two conditions must be checked before the formula is used. First, the equation must be in standard form, with coefficient 1 on y''. Divide by the leading coefficient first, because p is read from the divided equation. Dividing by x^2, for instance, changes the coefficient of y' to -\frac{2}{x} and gives a Wronskian that grows as a power of x. Second, the formula holds on any interval where p is continuous. When the leading coefficient vanishes at a point, work on an interval that avoids that point. The Wronskian can then blow up at the singular point.

Example: for y''-\frac{3}{x}y'+\dots=0 on x>0 with W(1)=2, we get W(x)=2\exp\left(\int_1^x\frac{3}{t}\,dt\right)=2x^3, so W(2)=16. For (x+1)y''+3y'+\dots=0 on x>-1, p=\frac{3}{x+1} and W(x)=\frac{C}{(x+1)^3}. With W(0)=4 we get C=4 and W(1)=\frac{1}{2}. The factor (x+1)^3 belongs in the denominator, since -\int p\,dx=-3\ln(x+1).

Euler–Cauchy equation

An Euler–Cauchy equation has the form x^2y''+axy'+by=0 for x>0. The power of x in each coefficient equals the order of the derivative it multiplies. Trying y=x^m gives the auxiliary equation m(m-1)+am+b=0. Distinct real roots m_1\neq m_2 give y=Ax^{m_1}+Bx^{m_2}. A repeated root m gives y=x^m(A+B\ln x). Complex roots \alpha\pm i\beta give x^{\alpha}\cos(\beta\ln x) and x^{\alpha}\sin(\beta\ln x).

The operator form explains why this works. Let t=\ln x and D=\frac{d}{dt}. Then x\frac{d}{dx}=D and x^2\frac{d^2}{dx^2}=D(D-1)=D^2-D, so the equation has constant coefficients in t. The most common error is to write x^2\frac{d^2y}{dx^2} as D^2y. The -D term must be kept. Matching D^2+(a-1)D+b against a target operator is therefore the correct procedure. For example, (D-2)^2y=0 corresponds to D^2-4D+4, and comparing constant coefficients fixes the constant in x^2y''-3xy'+(\text{const})y=0.

Another standard route is the substitution t=\ln x. An equation such as x^2y''+xy'-9y=0 becomes y_{tt}-9y=0, whose solutions are e^{3t}=x^3 and e^{-3t}=x^{-3}. Thus y=Ax^3+Bx^{-3}. Initial data at x=1 then fix A and B, and differentiating y gives y' at any other point. Keep in mind that \frac{d}{dx} of x^{-3} is -3x^{-4}, not a power of t.

ODE reduction by substitution

Changing the dependent variable can simplify an ODE. Put y(x)=u(x)v(x) with u nonzero. Then y'=u'v+uv' and y''=u''v+2u'v'+uv''. For the standard form y''+py'+qy=0, the coefficient of v' in the new equation becomes p+2\frac{u'}{u}, after the whole equation is divided by u. Choosing \frac{u'}{u}=-\frac{p}{2}, that is u=\exp\left(-\frac12\int p\,dx\right), removes the first-derivative term.