x^2y' + xy = x + 1

Question

find the general solution of the given differential equation. Give the largest interval I over which the general solution is defined. Determine whether there are any transient terms in the general solution.

Answer 1

x²y'+xy=x+1; divide through by x:

xy'+y=(x+1)/x=1+1/x,

(d/dx)(xy)=1+1/x,

xy=∫(1+1/x)dx=x+ln|x|+C, where C is the constant of integration.

y=1+(ln|x|+C)/x which can be written y=1+ln|ax|/x where a is a constant (a=e^C).

Note that x cannot be zero, so the y-axis is the vertical asymptote. This is a discontinuity. y=1 is the horizontal asymptote.

The largest interval I=(-∞,0) or (0,∞) which have the same size. The transient term is ln|ax|/x which approaches zero as x→∞, and the graph is levelling out to y=1.