What is the continuous solution to to the differential equation (x^2+5)y'+2xy=f(x), y(1)=1 where f(x)=0 when x<0, f(x)=x when 0<x<1 and f(x)=1 when x>1 ?

1 Answer
May 24, 2018
  • #y={( 11/(2(x^2+5)) " " x lt 0 ),( (x^2 + 11)/(2(x^2+5)) " " 0 lt x lt 1),( (x + 5)/(x^2+5) " " x gt 1) :}#

Explanation:

This DE is exact on the LHS:

  • #(x^2+5)y'+2xy=f(x) implies color(red)(((x^2+5)y)^' = f(x))#

Function is:

  • #f(x)={(0 " " x lt 0 ),( x " " 0 lt x lt 1),( 1 " " x gt 1) :}#

IV is:

  • #y(1) = 1#

Work backwards:

For #x gt 1#:

#((x^2+5)y)^' = 1#

#(x^2+5)y = x + C#

And

  • #y(1) =1 implies C = 5#

#implies y = (x + 5)/(x^2+5)#

For #0 lt x lt 1#:

#((x^2+5)y)^' = x#

#(x^2+5)y = x^2/2 + C#

With the same IV, #qquad y(1) =1#:

  • # implies C = 11/2#

#implies y = (x^2 + 11)/(2(x^2+5))#

For #x lt 0#:

#((x^2+5)y)^' = 0#

#(x^2+5)y = C#

And

  • #y(0) =11/10 implies C = 11/2#

#implies y = 11/(2(x^2+5))#

Solution is:

  • #y={( 11/(2(x^2+5)) " " x lt 0 ),( (x^2 + 11)/(2(x^2+5)) " " 0 lt x lt 1),( (x + 5)/(x^2+5) " " x gt 1) :}#