Mathematics: Find the general solution of x 1 = 3x1 - x2 + et, x 2 = x1.

Find the general solution of x 1 = 3x1 - x2 + - id13121501263, ramentome7542

Ответ:

drainy0uandthefish

15.03.2022

Note that if ${x_2}'=x_1$ , then ${x_2}''={x_1}'$ , and so we can collapse the system of ODEs into a linear ODE:

${x_2}''=3{x_2}'-x_2+e^t$
${x_2}''-3{x_2}'+x_2=e^t$

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

$r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0$

so that the characteristic solution is

${x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}$

Now let's suppose the particular solution is ${x_2}_p=ae^t$ . Then

${x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t$

and so

$ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1$

Thus the general solution for x_2

is

$x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t$

and you can find the solution x_1

by simply differentiating x_2

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Ответ:

thomisonariel9800

18.07.2021

Mathematics

Set it up like this
54.390
-36.245
=18.145
Borrow from the 9 so the 0 becomes 10 and the 9 now becomes 8. Subtract 10-5=5 then 8-4=4 the. 3-2=1 and we borrow from the 5 so the 4 becomes 14, now you subtract 14-6= 8 and the 5 is now 4 so 4-3=1

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Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.

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