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
And
#y(1) =1 implies C = 5#
For
With the same IV,
# implies C = 11/2#
For
And
#y(0) =11/10 implies C = 11/2#
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) :}#