# Find the differential equation given the general solution

Given that $x(t)=(c_1+c_2 t + c_3 t^2)e^t$ is the general solution to a differential equation, how do you work backwards to find the differential equation?

• What kind of differential equation it must satisfies? Linear, non-linear, first order,..? You can just differentiate it and get tons of differential equations for it, for example, differentiate $x$ and calculate $x'+x=?$. This is an differential equation. – Tomás Sep 5 '14 at 13:05
• You can differentiate both sides of the equations until you've deal with all C and rephrase values at each steps as iterations of the derivation of y. (y,y',y''.....y^n). – 123 Sep 5 '14 at 13:25

Say $f(t) = c_1 + c_2t + c_3t^2$
the given general solution is $$x(t) = fe^t$$ Since you have $3$ arbitrary constants, the required DE must be of order $3$. So you need to differentiate exactly 3 times : \begin{align} x' &= (f'+f)e^t \\ x''&=(f''+2f'+f)e^t \\x'''&=(f''' + 3f''+3f'+f)e^t\end{align}

Its trivial to eyeball the required DE : $x'''-3x''+3x'-x=f'''e^t = 0$

The general solution $$x(t) = (c_1 + c_2 t + c_3 t^2)\, e^t$$ has three parameters $c_i$, so one would need three integrations to reintroduce them from a differential equation, or three differentiations to eleminate them: \begin{align} x(t) &= (c_1 + c_2 t + c_3 t^2)\, e^t \Rightarrow \\ \dot{x}(t) &= (c_2 + 2\, c_3 t)\, e^t + x(t) \Rightarrow \\ \ddot{x}(t) &= 2\, c_3 \, e^t + 2 \dot{x}(t) - x(t) \Rightarrow \\ \dddot{x}(t) &= 2\, c_3 \, e^t + 2 \ddot{x}(t) - \dot{x}(t) \end{align} This would give $$\dddot{x}(t) - \ddot{x}(t) = 2\ddot{x}(t)-3\dot{x}(t)+x(t)$$ or $$\dddot{x}(t) - 3\ddot{x}(t) + 3\dot{x}(t) - x(t) = 0$$

Check: I am too lazy to solve this manually, so I used the machine: Wolfram Alpha solution

Given a homogeneous linear differential equation of order $n$, we can get the solutions by writing the characteristic polynomial and equating to zero.

$c_nx^{(n)}(t)+c_{n-1}x^{(n-1)}(t)+...+c_{1}x^{(1)}(t)+c_0x(t)=0$

$x(t)=\sum_iA_ie^{\lambda_i t}$, where $\lambda_i$ is a $\lambda$ such that solves the characteristic polynomial:

$c_n\lambda^{n}+c_{n-1}\lambda^{n-1}+...+c_1\lambda^{1}+c_0=0$

However, if some $\lambda_i$ is a solution of the characteristic polynomial with a multiplicity $k>1$, its contribution to the general $x(t)$ is not $Ae^{\lambda_i t}$, but $\sum_{j=0}^{k-1}A_jt^{j}e^{\lambda_i t}$. For example, if the characteristic polynomial is $(\lambda-2)^2$ then the solution is not $Ae^{2t} + Be^{2t}$, but $Ae^{2t} + Bte^{2t}$.

This is exactly our case. We can see the solution is of the form $\sum_{j=0}^{k-1}A_jt^{j}e^{\lambda_i t}$ where $\lambda_i=1$, $k-1=2\rightarrow k=3$.

So the characteristic polynomial is $(\lambda-1)^3=\lambda^3-3\lambda^2+3\lambda-1$.

From this we can conclude that the differential equation is $x^{(3)}(t)-3x^{(2)}(t)+3x^{(1)}(t)-x(t)=0$.