After some algebraic simplification, I got the ODE: $\ddot x(t)+\sqrt {(\dot x(t)+x(t))^2}+k x(t)=0$ I interpreted this equation as: $\ddot x(t)+|{(\dot x(t)+x(t))}|+k x(t)=0$ I have some problem to solve it. Could you give me some hint please? Thanks.
Differential equation with absolute value
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ordinary-differential-equations
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0You may use Godisemo's hint and then it becomes a 2nd ordered linear ODE with constant coefficients. – 2012-03-16
1 Answers
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Your equation is equivalent to
$\ddot x(t)\pm\dot x(t)\pm x(t)+kx(t)=0$
so you have to solve the equations
$\ddot x(t)+\dot x(t)+(1+k)x(t)=0 \qquad \ddot x(t)-\dot x(t)-(1-k)x(t)=0.$
The solutions of these equations can be obtained by solving the characteristics equations
$\lambda^2\pm\lambda+(\pm 1+k)=0$
giving
$\lambda_{1,2}=\frac{\mp 1\pm\sqrt{1-4(\pm 1+k))}}{2}$
and depending on the value of $k$ you will get different sets of solutions.