Question

Solve the given initial-value problem. give the largest interval i over which the solution is defined. x dy dx + y = 6x + 1, y(1) = 8

Answer 1

$x\dfrac{\mathrm dy}{\mathrm dx}+y=6x+1$
$\dfrac{\mathrm d}{\mathrm dx}(xy)=6x+1$
$xy=\displaystyle\int(6x+1)\,\mathrm dx$
$xy=3x^2+x+C$

Given that $y(1)=8$, we have

$8=3+1+C\implies C=4$

so that the particular solution is

$xy=3x^2+x+4$
$y=3x+1+\dfrac4x$

which is valid as long as $x\neq0$, which in turn suggests the largest interval over which the solution may be valid is $(0,\infty)$.