(a)
We have to solve the initial value problem given by:
x'=10y
y'=-10x
with initial conditions:
x(0)=3
y(0)=4
Now we can write the above problem in matrix form as shown:
[[x'],[y']]=[[0,10],[-10,0]][[x],[y]]
So let A=[[0,10],[-10,0]]
Now let us write the characteristic equation i.e.
|A-lambda I |=0
|[-lambda,10],[-10,-lambda]|=0
lambda^2+100=0
rArr lambda = +-10i
Now we have to find the eigen vectors corresponding to the one of the eigen values obtained above.
For lambda_1=10i
We have,
[[-10i,10],[-10,-10i]][[v_1],[v_2]]=[[0],[0]]
i.e.
-10i v_1+10v_2=0
-iv_1+v_2=0 rArr v_2=iv_1
or,
-10v_1-10iv_2=0
v_1+iv_2=0
i.e. -iv_1+v_2=0 rArr v_2=iv_1
So we have the eigen vector as:
eta_1=[[v_1],[v_2]]=[[1],[i]] =[[1],[0]]+[[0],[1]]i
when v_1=1
So now we can write the solution as:
Since we have complex conjugate eigen values of the form mu+-lambda i and suppose eta = a+bi is the eigen vector,
our solution will be of the form:
[[x],[y]]=C_1 e^{mu t}(a cos(lambda t)-bsin( lambda t))+C_2e^{mu t}(asin(lambda t)+bcos(lambda t))
i.e. [[x],[y]]=C_1e^{0 t}([[1],[0]]cos(10 t)-[[0],[1]]sin(10t))+C_2e^{0 t}([[1],[0]]sin(10 t)+[[0],[1]]cos(10 t))
=C_1[[cos(10t)],[-sin(10t)]]+C_2[[sin(10t)],[cos(10t)]]
Now applying the initial conditions we have,
[[x(0)],[y(0)]]=[[3],[4]]=C_1[[1],[0]]+C_2[[0],[1]]
i.e.
C_1=3 and C_2=4
Hence we have the final solution as:
[[x(t)],[y(t)]]=3[[cos(10t)],[-sin(10t)]]+4[[sin(10t)],[cos(10t)]]
i.e.
x(t)=3cos(10t)+4sin(10t) and,
y(t)=-3sin(10t)+4cos(10t)
(b)
Now we will sketch the graphs of the parametric equations x(t) and y(t)
The graph is of the shape of a circle with radius 5.
Location of initial conditions are also shown.
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