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You can derive two times the identity $y=C_1e^{-x} + C_2 x e^{-x}$, take the two derivatives for solving $C_1$ and $C_2$, and then replace these constants in the initial identity, that is:

var("C1, C2")
y = function("y")(x)
f = C1*e^(-x) + C2*x*e^(-x)
sol = solve([diff(y==f,x), diff(y==f,x,2)], C1, C2)
ode = (y-f==0).subs(sol).full_simplify()
show(ode)

The output is

$$y\left(x\right) + 2 \, \frac{\partial}{\partial x}y\left(x\right) + \frac{\partial^{2}}{(\partial x)^{2}}y\left(x\right) = 0$$

You can prettify it by using, for example, the following code inspired by @slelievre's answer to this question, i.e.:

y1 = function("y1", latex_name="y'")(x)
y2 = function("y2", latex_name="y''")(x)
ode_pretty = ode.subs({diff(y(x), x):y1, diff(y(x), x, x): y2})
show(ode_pretty)

which yields

$$y\left(x\right) + 2 \, y'\left(x\right) + y''\left(x\right) = 0$$