# Solution to $1+x+r\dot x -\ddot x=0$ when $x(0)=1$: combination of monomials with solutions to the characteristic equation?

Suppose $1+x+r\dot x -\ddot x=0$ when $x(t_h)=x(0)=1$.

Now the characteristic equation is $-h^2+rh+1=0$ and its roots are $h=\frac{-r\pm\sqrt{r^2+4}}{-2}$. So the solution is their linear combination:

$$x^*_1=C_1e^{\frac{-r+\sqrt{r^2+4}}{-2}t}+C_2e^{\frac{-r-\sqrt{r^2+4}}{-2}t}=C_1e^{\frac{r-\sqrt{r^2+4}}{2}t}+C_2e^{\frac{r+\sqrt{r^2+4}}{2}t}$$

and by $x(0)=1=C_1+C_2$. This is the way I have learnt it from this tutorial here, the equation one with the positive determinant under the second order-differential equations.

$$x^*_2=2e^{0.5(r+\sqrt{4+r^2})t}+C_1e^{0.5(r-\sqrt{4+r^2})t}-C_1e^{0.5(r+\sqrt{4+r^2})t}-1$$

and I can see that $x^*_2$ satisfy the condition $x(0)=2+1-1-1=1$ so I cannot see anything wrong with the last method but I would like to understand how the first method would work.

How would the first method work to get $x_1^*$?

• Thank you! Got it now clear :)
– hhh
Commented Apr 29, 2014 at 10:19

If you do not use the boundary condition $x(0)=1$, the solution of the differential equation is $$x(t)=c_1 e^{\frac{1}{2} \left(r-\sqrt{r^2+4}\right) t}+c_2 e^{\frac{1}{2} \left(\sqrt{r^2+4}+r\right) t}-1$$ So $x(0)=c_1+c_2-1=1$.

Is this clearer ? You just forgot the fact that $x(t)=-1$ is a solution.

• Where/how did you get the particular solution? What does this $x(t)=-1$ mean? I understand the mixing of homogenous situation to this non-homegenous situation, +1 for it! It is the last part of the article here. My thinking: $1+x+r\dot x -\ddot x=0$ where $-1$ is the solution to $x+r\dot x -\ddot x$ i.e. $x+r\dot x -\ddot x=-1$? So general solution is just the sum of homogenous plus the particular solution.
– hhh
Commented Apr 29, 2014 at 10:02
• How is this $2e^{0.5(r+\sqrt{4+r^2})t}+C_1e^{0.5(r-\sqrt{4+r^2})t}-C_1e^{0.5(r+\sqrt{4+r^2})t}$ the homogenous solution? I can understand your ideas but not yet the idea here.
– hhh
Commented Apr 29, 2014 at 10:09
• You have $x(0)=c_1+c_2-1=1$ which means that $c_2=2-c1$. Now replace in the expression I wrote. Commented Apr 29, 2014 at 10:15
• Same time! $2e^{0.5(r+\sqrt{4+r^2})t}+C_1e^{0.5(r-\sqrt{4+r^2})t}-C_1e^{0.5(r+\sqrt{4+r^2})‌​t}=(2-C_1)e^{0.5(r+\sqrt{4+r^2})t}+C_1e^{0.5(r-\sqrt{4+r^2})t}=C_2e^{0.5(r+\sqrt{4+r^2})t}+C_1e^{0.5(r-\sqrt{4+r^2})t}$. Bingo, Thank you! :D
– hhh
Commented Apr 29, 2014 at 10:17
• @hhh. You are welcome ! Commented Apr 29, 2014 at 10:22

$x_1^*$ is a solution of the homogeneous equation $x+r\,\dot x-\ddot x=0$, but not of the complete equation $1+x+r\,\dot x-\ddot x=0$

• We reacted inside the same second ! Cheers. Commented Apr 29, 2014 at 9:54
• Spot-on, messed up homogenous and non-homegenous cases +1 for it. I can understand so far but not yet $2e^{0.5(r+\sqrt{4+r^2})t}+C_1e^{0.5(r-\sqrt{4+r^2})t}-C_1e^{0.5(r+\sqrt{4+r^2})t}$ in $x^*_2=2e^{0.5(r+\sqrt{4+r^2})t}+C_1e^{0.5(r-\sqrt{4+r^2})t}-C_1e^{0.5(r+\sqrt{4+r^2})t}-1$. I would write it as Claude i.e. $x(t)=c_1 e^{\frac{1}{2} \left(r-\sqrt{r^2+4}\right) t}+c_2 e^{\frac{1}{2} \left(\sqrt{r^2+4}+r\right) t}-1$.
– hhh
Commented Apr 29, 2014 at 10:06