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(\mathrm dy)/(\mathrm dx)+y=6x+1`

`(\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+\frac4x`

which is valid as long as
`x\\eq0`, which in turn suggests the largest interval over which the solution may be valid is
`(0,\infty)`.