ODE Theory: General Solution to y'' + 4y' + 4y = 0

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SUMMARY

The general solution to the differential equation y'' + 4y' + 4y = 0 is y = c1*exp(-2t) + c2*t*exp(-2t). The characteristic equation yields a double root at -2, necessitating the use of reduction of order to find the second solution. The confusion arises from the incorrect assumption that the solution can be expressed as c1*e^(-2t) + c2*e^(-2t), which does not account for the multiplicity of the root. The correct approach involves using the form y = u(t)e^(-2t) to derive the general solution accurately.

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Homework Statement



We have y'' + 4y' + 4y = 0 ; find the general solution.

Homework Equations



Reduction of Order.

The Attempt at a Solution



So when determining the roots of the characteristic equation, -2 was a double root, and therefore we can't simply have c1e-2t + c2e-2t. So I thought I would use reduction of order to get a second equation. However in the solution, they just left it c1e-2t + c2e-2t and I'm wondering if what I was taught to do in the case of non distinct roots was wrong, or if the solution is wrong.
 
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The solution appears to be wrong.

y = c1*exp(-2t) + c2 * t * exp(-2t)
 
Please show exactly what you did in your attempted reduction of order. When I try a solution of the form [itex]y= u(t)e^{-2t}[/itex], I get [itex]u(t)= A+ Bt[/itex] giving [itex]y= Ae^{-2t}+ Bte^{-2t}[/itex] as general solution.
 

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