The equation y'' + 3 x y' + y = 0

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Ordinary differential equations · Ordinary differential equations

Conditions of Validity, Ch Linear ordinary differential equations: existence, the Wronskian, and a kernel in disguise

Problem

The equation y′′+3xy′+y=0,x>0,y'' + \frac{3}{x}y' + y = 0, \qquad x \gt 0, has coefficients continuous on (0,∞)(0,\infty), so it has two independent solutions y1,y2y_1,y_2 there with Wronskian W(x)=y1y2′−y2y1′W(x)=y_1y_2'-y_2y_1'. This is a purely mathematical statement and is dimensionless throughout. ((a)) Write the equation as the first-order system y′=A(x)y\mathbf{y}'=\mathsf{A}(x)\mathbf{y} and give tr A(x)\mathrm{tr}\,\mathsf{A}(x). ((b)) Use Liouville's formula W′=(tr A)WW'=(\mathrm{tr}\,\mathsf{A})W to find the power nn such that W(x)∝xnW(x)\propto x^n. ((c)) Confirm the result using Abel's identity W(x)=W(x0)exp⁡ ⁣(−∫x0xa1(s) ds)W(x)=W(x_0)\exp\!\left(-\int_{x_0}^{x}a_1(s)\,\mathrm{d}s\right) with a1(x)=3/xa_1(x)=3/x.

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