Mathematics: Find the general solution to the homogeneous differential equation dt2d2y+18dtdy+81y=0 The solution can be written in the form y=C1f1(t)+C2f2(t) with f1(t)= and f2(t)=

Step-by-step explanation:27 meter!!...

Find the general solution to the homogeneous differential

Ответ:

destinyd10189

19.01.2022

Mathematics

$y(t) = C_1 e^{-9t}+C_2te^{-9t}$

Step-by-step explanation:

Consider the differential equation $\frac{d^2y}{dt^2} + 18 \frac{dy}{dt}+81y=0$ . Let us propose a solution of the form $y = A e^{rt}$ where A is a non-zero constant. Then, the derivatives are calculated as follows: $\frac{dy}{dt} = Ar e^{rt},\frac{d^2y}{dt^2}= Ar^2e^{rt}$ . Replacing those functions in the original equation, we obtain the following.

$Ar^2e^{rt}+18Are^{rt}+81Ae^{rt}=0= Ae^{rt}(r^2+18r+81)=0$

Given that A is non-zero and that the exponential function is non-zero, the following equation must be hold

r^2+18r+81 = 0=(r+9)^2

Then, the only solution is r=-9. Then, one of the solutions is the function [text]y_1(t)=C_1 e^{-9t}[/tex]. Since the differential equation is a second degree equation, another solution must be given. Note that the second solution must be linear independent of the first found solution. This is achieved by multiplying the first solution by t i.e $y_2(t) = C_2 te^{-9t}$ . Then, the complete solution is of the form $y(t) = C_1 e^{-9t}+C_2te^{-9t}$

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Find the general solution to the homogeneous differential equation dt2d2y+18dtdy+81y=0 The solution can be written in the form y=C1f1(t)+C2f2(t) with f1(t)= and f2(t)=

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