Differential equation with the solution of $(1+ax/2)\exp(-ax)$

Question

Is there any linear differential equation which has following solution

$$y=(1+ax/2)\exp(-ax)$$ $a$ is constant.

Answer 1

Let $y = (1 + ax/2)e^{-ax}$ and $D = d/dx$. Note that, $$ 2ay = 2ae^{-ax} + a^2xe^{-ax} $$ $$ 2Dy = 2D\biggl(e^{-ax} + \dfrac{axe^{-ax}}{2}\biggr) = -2ae^{-ax} + ae^{-ax} -a^2xe^{-ax} = ae^{-ax} - 2ay \quad \Rightarrow $$ $$ 2D^2y = D(2Dy) = -a^2e^{-ax} - 2aDy = -a(ae^{-ax}) - 2aDy \quad \Rightarrow $$ $$ 2D^2y + 2aDy = -a(2Dy + 2ay) \quad \Rightarrow \quad (D^2 + 2aD + a^2)y = 0 \quad \Rightarrow \quad (D + a)^2y = 0 $$ or $$ y^{\prime \prime} + 2ay^{\prime} + a^2y = 0 $$ General solution: $y = C_1e^{-ax} + C_2xe^{-ax}$

Answer 2

For all differentiable functions $y=f(x)$, one can always obtains a linear ode by taking derivative on both sides. Say, $f(x)=(1+ax/2)\exp{(-ax)}$ has immediately ode $$y'(x)=f'(x)=-a(1+ax)\exp{(-ax)}/2$$

Answer 3

Here is an approach. Let

$$ y = (1+a\,x/2)e^{-ax} \longrightarrow (*) $$

Differentiating w.r.t. $x$ both sides we have

$$ y'= -a(1+ax/2)e^{-ax}+\frac{a}{2}e^{-ax} $$

Using $(*)$ we get

$$ y'=-y+\frac{a}{2}\frac{y}{(1+ax/2)} $$

$$ \implies y' = \left(\frac{a}{2}\frac{1}{(1+ax/2)}-1\right)y,\quad y(0)=1. $$

Added: I already introduced a technique for the question you are asking in my previous work. See for instance section 6, lemma 1 or see my book. Another approach is the annihilator techniques.

Answer 4

Hint: if you want a constant-coefficient homogeneous linear differential equation, you'll want the characteristic polynomial to have a double root at $-a$.