Classification of solutions of x αx x 0 1 α 2 there

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Classification of Solutions of x ′′ αx + x = 0 1. α = 2 . There is one real solution. This case is called critical damping since the solution r = 1 leads to exponential decay. The solution is x ( t ) = ( c 1 + c 2 t ) e t . 2. α < 2 . There are two real, negative solutions, r = μ, ν, μ,ν > 0 . The solution is x ( t ) = c 1 e μt + c 2 e νt . In this case we have what is called overdamped motion. There are no oscil- lations 3. 2 <α< 0 . There are two complex conjugate solutions r = α/ 2 ± with real part less than zero and β = 4 α 2 2 . The solution is x ( t ) = ( c 1 cos βt + c 2 sin βt ) e αt/ 2 . Since α< 0 , this consists of a decaying exponential times oscillations. This is often called an underdamped oscillation . 4. α = 0 . This leads to simple harmonic motion . 5. 0 < α < 2 . This is similar to the underdamped case, except α> 0 . The solutions are growing oscillations. 6. α = 2 . There is one real solution. The solution is x ( t ) = ( c 1 + c 2 t ) e t . It leads to unbounded growth in time. 7. For α> 2 . There are two real, positive solutions r = μ,ν> 0 . The solution is x ( t ) = c 1 e μt + c 2 e νt , which grows in time. For α< 0 the solutions are losing energy, so the solutions can oscillate with a diminishing amplitude. For α> 0 , there is a growth in the amplitude, which is not typical. Of course, there can be overdamped motion if the magnitude of α is too large. Example 2.9. Degenerate Node x = x y = 2 x y. (2.23) For this example, we write out the solutions. While it is a coupled system, only the second equation is coupled. There are two possible approaches. a. We could solve the first equation to find x ( t ) = c 1 e t . Inserting this solution into the second equation, we have y + y = 2 c 1 e t . This is a relatively simple linear first order equation for y = y ( t ) . The inte- grating factor is μ = e t . The solution is found as y ( t ) = ( c 2 2 c 1 t ) e t . b. Another method would be to proceed to rewrite this as a second order equation. Computing x ′′ does not get us very far. So, we look at
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38 2 Systems of Differential Equations -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 x(t) y(t) α =0.1 Fig. 2.14. Phase portrait for Example 2.8 with α = 0 . 1. This is an unstable focus, or spiral. y ′′ = 2 x y = 2 x y = 2 y y. (2.24) Therefore, y satisfies y ′′ + 2 y + y = 0 . The characteristic equation has one real root, r = 1 . So, we write y ( t ) = ( k 1 + k 2 t ) e t . This is a stable degenerate node. Combining this with the solution x ( t ) = c 1 e t , we can show that y ( t ) = ( c 2 2 c 1 t ) e t as before. In Figure 2.16 we see several orbits in this system. It differs from the stable node show in Figure 2.2 in that there is only one direction along which the orbits approach the origin instead of two. If one picks c 1 = 0, then x ( t ) = 0 and y ( t ) = c 2 e t . This leads to orbits running along the y -axis as seen in the figure.
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