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I'm trying to find the solution to the differential equation for a damped harmonic oscillator, i.e. $m\ddot{x}+c\dot{x}+kx=0$ but using that the damping force can be represented by the frictional force, $f_k$. This gives $$m\ddot{x}+kx+f_k=0 \hspace{3mm} \mbox{ for }\dot{x}>0 \\ m\ddot{x}+kx-f_k=0 \hspace{3mm} \mbox{ for }\dot{x}<0$$

I also have initial conditions $x_{t=0}=x_0$ and $v_{t=0}=0$.

A complementary solution to this equation is $x_c=A\cos({wt-\phi})$ where $A$ is a constant and a particular solution is $x_p=\mp\frac{f_k}{k}$ (negative for $\dot{x}>0$ and positive for $\dot{x}<0$).

Therefore we obtain the general solution $$x(t)=A\cos(wt-\phi)-\frac{f_k}{k} \hspace{3mm} \mbox{ for } \dot{x}>0\\ x(t)=A\cos(wt-\phi)+\frac{f_k}{k} \mbox{ for } \dot{x}<0$$

Unfortunately this doesn't make sense since the output values of this function don't get smaller with time.

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It seems that you are calling $c\dot{x}$ as $f_k$, solving the equation as if $f_k$ was a parameter, then trying to substitute this constant somehow back. This is not a way to go.

Also there is no point it single out cases $\dot{x}>0 $ and $\dot{x} < 0.$ Both are covered by $c \dot{x}=0$ (assuming by "friction" you mean really something which is trying to stop a particle for $c>0$). In your case by changing sign the $\dot{x}$ term is slowing down or speeding up the particle based on the sign of variable. Actually this does not matted as you do not solve the equation properly.

To solve it properly you should consult online material or any book. For example this note from the MIT OCW Differential Equations coarse.

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  • $\begingroup$ Yea I agree with what you've said but it's just odd since this is how my professor asked us to solve it. $\endgroup$
    – Craig
    Commented Feb 18, 2016 at 0:39

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