Question

consider the second order linear differential equation: y −2ry +r²y=0 Notice that the auxillary polynomial has a root of multiplicity 2 . In this case, show that y1 (t)=eπ and y₂(t)=teʳᵗ are both solutions to the differential equation above.

Answer 1

To confirm y1(t) = e^(rt) and y2(t) = t*e^(rt) as solutions to the differential equation y - 2ry' + r^2y = 0, we differentiate them and substitute into the equation, finding they indeed satisfy the equation and simplify to zero, thus verifying their validity as solutions.

Step-by-step explanation:

To show that y1(t) = ert and y2(t) = tert are solutions to the given differential equation y - 2ry' + r2y = 0, we start by assuming they are solutions and substituting them into the equation.

For y1(t) = ert, differentiation gives y1' = rert and y1'' = r2ert. Substituting these into the differential equation yields:
ert - 2r(rert) + r2(ert) = 0, which simplifies to zero, verifying that y1(t) is a solution.

For y2(t) = tert, differentiation gives y2' = ert + rtert and y2'' = 2rert + r2tert. Substituting these into the differential equation yields:
(tert) - 2r(ert + rtert) + r2(tert) = 0, which also simplifies to zero, verifying that y2(t) is a solution.