Solve \( y'(x)=1+2 x\) for \(y(x)\) with \(y(0)=3\)
import sympy x = sympy.symbols('x') y = sympy.Function('y') ode = sympy.Eq(sympy.Derivative(y(x),x),1+2*x) sol = sympy.dsolve(ode,y(x),ics={y(0):3}) # Eq(y(x), x**2 + x + 3) sympy.checkodesol(ode,sol) # (True, 0)