# Motion of particle in uniform magnetic field with friction propositional to velocity [closed]

A positively charged particlewith charge "q" is at $$(0,0,0)$$.There is a uniform magnetic field $$\vec B= -B \hat k$$. The particle is given velocity $$\vec u= u \hat i$$. A resistive friction acts on it which is proportional to is velocity $$(\vec F_{friction} = - k \vec v)$$. Find the equation of the trajectory of its motion on the trajectory of its motion

I considered its velocity at any time t to be $$\vec v = v_x \hat i + v_y \hat j$$ Now Force on it is $$\vec F= q(v_x \hat i + v_y \hat j)×(-B \hat k) - k( v_x \hat i + v_y \hat j)$$

So $$\vec F= q(-v_y B - k v_x) \hat i+ q(v_x B - k v_y) \hat j$$

So know I get 2 differential equations $$m\frac{d V_x}{dt} = q(-v_y B - k v_x )$$ And $$\frac{dV_y}{dt} = q(v_x B - k v_y )$$

Now what... How to solve these ??

[ What I originally planned was to seperate these differential equations to get velocity of $$v_x and v_y$$ and then integrate it {Like we do it in the question with cycloid, where there is an $$\vec E$$ field perpendicular to $$\vec B$$ field }]

(I'm in highschool so , maybe this might be out of my reach buy since I was presented this problem, I assume there surely must be some clever trick to solve it, or any other shorter approach. Please help :)

You should seek a solution of the form $$v_x = A\exp(l\cdot t)$$ and $$v_y = B\exp(l\cdot t)$$. After you put it into the equations, you will received two linear algebraic equations for $$A$$ and $$B$$ ($$l$$ is a parameter). Choose a parameter so that there are non-trivial solutions for $$A,\,B$$. And solve this two linear equations with the parameter $$l$$ (maybe there will be two suitable values for $$l$$ and you should solve with each one separately).