# Thread: A DE I fell over in a physics problem

1. ## A DE I fell over in a physics problem

Hello!
I had to solve a physics problem (related to gases) and I fell over a DE that I don't know how to solve. (I didn't see its type in Chris' tutorial).
I must find $x(t)$ such that $\frac{d^2x}{dt^2}=Gx(t)+K$, where G and K are constants.
I probably made a mistake, because I don't know if we can solve it algebraically and it should describe a harmonic motion.
I appreciate any help in solving it. Thank you.

2. Originally Posted by arbolis
Hello!
I had to solve a physics problem (related to gases) and I fell over a DE that I don't know how to solve. (I didn't see its type in Chris' tutorial).
I must find $x(t)$ such that $\frac{d^2x}{dt^2}=Gx(t)+K$, where G and K are constants.
I probably made a mistake, because I don't know if we can solve it algebraically and it should describe a harmonic motion.
I appreciate any help in solving it. Thank you.
it's a simple non-homogeneous linear differential equation with constant coefficients. if $G=0,$ then integrating twice w.r.t. $t$ will give you $x(t).$ if $G \neq 0,$ then a particular solution is $x_p=\frac{-K}{G}.$

so you only need to solve $x''=Gx,$ which has characteristic polynomial $r^2=G.$ so now the answer depends on whether $G > 0$ or $G<0.$ anyway, i think you can take it from here.

3. Originally Posted by arbolis
Hello!
I had to solve a physics problem (related to gases) and I fell over a DE that I don't know how to solve. (I didn't see its type in Chris' tutorial).
I must find $x(t)$ such that $\frac{d^2x}{dt^2}=Gx(t)+K$, where G and K are constants.
I probably made a mistake, because I don't know if we can solve it algebraically and it should describe a harmonic motion.
I appreciate any help in solving it. Thank you.

we are trying to solve

$x''-Gx=K$

first we would solve the homogenious equation

$x''-Gx=0$

If $G>0$

then we get

$m^2-G=0 \iff m \pm \sqrt{G}$

so $x_c=c_1e^{t\sqrt{G}}+c_2e^{-t\sqrt{G}}$

and $x_p=Ax^2+Bx+C \implies x_p''=2A$

$2A+G(Ax^2+Bx+C)=K \implies A=0,B=0,C=\frac{K}{G}$

So we get $x(t)=x_c+x_p=c_1e^{t\sqrt{G}}+c_2e^{-t\sqrt{G}}+\frac{K}{G}$

If $G<0$ you use a the same proceedure.

4. Thank you both.
NCA, I don't understand why you wrote
so you only need to solve
. Indeed from it I recognize that the solutions are $x(t)=A\cos (\omega t +\phi)$ but I don't understand why $K$ can be considered as $0$. Also I don't really know how to derive that the solutions are under the form $x(t)=A\cos (\omega t +\phi)$ but I'm sure I can find the solution in Chris' tutorial.
By the way $K>0$. While $G$ can be anything.
TheEmptySet, I didn't follow you when you wrote " $m$" and " $x_p$" and " $x_c$". Could you explain a bit more? Thanks in advance.

5. assuming the K is a constant, the particulate solution would just be x=A

correct?

6. Originally Posted by arbolis
Thank you both.
NCA, I don't understand why you wrote . Indeed from it I recognize that the solutions are $x(t)=A\cos (\omega t +\phi)$ but I don't understand why $K$ can be considered as $0$. Also I don't really know how to derive that the solutions are under the form $x(t)=A\cos (\omega t +\phi)$ but I'm sure I can find the solution in Chris' tutorial.
By the way $K>0$. While $G$ can be anything.
TheEmptySet, I didn't follow you when you wrote " $m$" and " $x_p$" and " $x_c$". Could you explain a bit more? Thanks in advance.