2nd Order ODE: General Solution

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2nd Order ODE: General Solution (Real)

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randRangeExclude(-5,5, [-1,0,1]) randRangeExclude(-5,5, [-1,0,1,-L1,1-L1,L1-1]) L1+L2 L1*L2 T*T-4*D

Consider the ODE y''(t) - T\cdot y'(t) +D\cdot y(t) = 0.

Determine the general solution y with y(t) and constants A and B, where A should precede B in the sense that A is associated to the smaller eigenvalue and B to the larger eigenvalue.

(A+B*t)e^{L1*t}

(A*t+B)e^{L1*t}

Ae^{L1*t} + Be^{L2*t}

Ae^{L2*t} + Be^{L1*t}

{y(t)} =

First use the characteristic equation \lambda^2 - T\cdot \lambda +D = 0 to get the general solution of the ODE.

The two (distinct) roots of the quadratic equation are \lambda_1 = L1 and \lambda_2 = L2.

Thus, the general solution is y(t) =A \cdot e^{L1 t} + B \cdot e^{L2 t} with constants A and B.

There is a double root of the quadratic equation \lambda_{1,2} = L1.

Thus, the general solution is y(t) =(A+Bt)e^{L1t} with A and B.