Your differential equation being linear first order, its solution involves one integration constant, which cannot (in general) satisfy two boundary conditions.
If you need a verification :
sage: %cpaste
Pasting code; enter '--' alone on the line to stop or use Ctrl-D.
:
:y = function("y")
:de = y(x)+y(x).diff(x) == x^2
:sol = desolve(de, y(x)) ; sol
:# Declare the integration constant
:Csts = [str(u) for u in sol.variables() if str(u) not in globals()]
:if len(Csts)>0 : var(Csts)
:# Check solution
:chk = bool(de.substitute_function(y, sol.function(x)))
:# Satisfy the first boundary condition:
:C1=solve(sol(x=1)==1, _C) ; C1
:# Satisfy the second boundary condition:
:C2 = solve(sol.diff(x)(x=1)==1, _C) ; C2
:
:--
((x^2 - 2*x + 2)*e^x + _C)*e^(-x)
[_C == 0]
[_C == -e]
your boundary conditions are mutually incompatible...
HTH,
EDIT : The same principle applies ; and since your DE is linear second order, two integration constants appear, which may fullfill two boundary conditions : :
sage: y=function("y")
sage: de=y(x)+y(x).diff(x,2)==x^2
sage: sol=desolve(de,y(x)) ; sol
x^2 + _K2*cos(x) + _K1*sin(x) - 2
sage: Csts = [str(u) for u in sol.variables() if str(u) not in globals()] ; Csts
['_K1', '_K2']
sage: if len(Csts)>0 : var(Csts)
(_K1, _K2)
# Solving for simultaneous boundary conditions :
sage: solve([sol(x=1)==1, sol.diff(x)(x=2)==1],(_K1, _K2))
[[_K1 == -(3*cos(1) - 2*sin(2))/(cos(2)*cos(1) + sin(2)*sin(1)), _K2 == (2*cos(2) + 3*sin(1))/(cos(2)*cos(1) + sin(2)*sin(1))]]
Note : Now, your question strongly smells of an (initialy mistyped) homework. Are you sure that, by taking the ask.sagemath.org shortcut, you aren't missing the point of said homework ?
