How to solve \(\displaystyle\text{f{x}''(t)=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}\textbf{x}(t}{f}{\left\lbrace{x}\right\rbrace}{''}{\left({t}\right)}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{1}&{0}\backslash{0}&-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\text{f{x}(t}{f}{\left\lbrace{x}\right\rbrace}{\left({t}\right)}\) What I'm thinking is

amonitas3zeb

amonitas3zeb

Answered question

2022-03-31

How to solve Kx''(t)=-100-1x(t)
What I'm thinking is to consider the first order version
x'(t)=-100-1x(t)
which I know how to solve, the solution is
x(t)=ce-te-t
How do I use this to solve the second-order equation?

Answer & Explanation

Boehm98wy

Boehm98wy

Beginner2022-04-01Added 18 answers

Step 1
Let y(t)=(x(t),x(t)). Then we have
y'(t)=0I2-I20y(t)
Then once you have the solution y(t)=eAty(0), you can get a solution for x.
Step 2
This is the general strategy. However, in your case, we have a simpler problem because the matrix is diagonal. Here,
x1''(t)x2''(t)=-x1(t)-x2(t)
and hence x1 and x2 are solutions to the one-dimensional equation u =u. So xi=aicos(t)+bisin(t) for i=1,2.

zevillageobau

zevillageobau

Beginner2022-04-02Added 13 answers

Step 1
Notice that the matrix you have at the beginning is the unit matrix of dimention 2, I2. Then you have (x(t),y(t))=(x(t),y(t)) so xt)+x(t)=0. The characteristic polynomial for this ODE is r2=1 which means r1=i and r2=i hence x(t)=c1eit+c2eit and hence your solution (by doing the same thing for y(t)) is
Step 2
(x(t),y(t))=(c1eit+c2eit,c3eit+c4eit).

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