Answer:
[tex]x=-cos(t)+2sin(t)[/tex]
Step-by-step explanation:
The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:
A differential equation of the form:
[tex]a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0[/tex]
Will have a characteristic equation of the form:
[tex]a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0[/tex]
Where solutions [tex]r_1,r_2...,r_n[/tex] are the roots from which the general solution can be found.
For real roots the solution is given by:
[tex]y(t)=c_1e^{r_1t} +c_2e^{r_2t}[/tex]
For real repeated roots the solution is given by:
[tex]y(t)=c_1e^{rt} +c_2te^{rt}[/tex]
For complex roots the solution is given by:
[tex]y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)[/tex]
Where:
[tex]r_1_,_2=\lambda \pm \mu i[/tex]
Let's find the solution for [tex]x''+x=0[/tex] using the previous information:
The characteristic equation is:
[tex]r^{2} +1=0[/tex]
So, the roots are given by:
[tex]r_1_,_2=0\pm \sqrt{-1} =\pm i[/tex]
Therefore, the solution is:
[tex]x(t)=c_1cos(t)+c_2sin(t)[/tex]
As you can see, is the same solution provided by the problem.
Moving on, let's find the derivative of [tex]x(t)[/tex] in order to find the constants [tex]c_1[/tex] and [tex]c_2[/tex]:
[tex]x'(t)=-c_1sin(t)+c_2cos(t)[/tex]
Evaluating the initial conditions:
[tex]x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1[/tex]
And
[tex]x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2[/tex]
Now we have found the value of the constants, the solution of the second-order IVP is:
[tex]x=-cos(t)+2sin(t)[/tex]
