User:EGM6321.f10.team6.yoshioka/HW3

Problem 5
The equation 5.2 can be re-arranged in the following form:

It is clear that this equation complies with the first condition of exactness.

So to make this equation exact we will apply the Euler Integrating Factor Method.

It has been previously shown that in order to meet the second condition of exactness we can multiply equation 5.3 by a function h to obtain the following:

Now the second condition of exactness may be applied, where x is the dependent variable and t is the independent variable.

Now for the special case where $${{h}_{x}}$$ equals zero we obtain the following:

Which may be rearranged to obtain:

We can now solve for h as a function of t.

Rewriting equation 5.3 we obtain:

We should recognize equation 5.10 as the derivative of (hx):

Therefore, to find the solution we integrate both sides in the interval $${{t}_{o}}\le \tau \le t$$.

Equation 5.12 yields:

Rearranging terms in equation 5.13 we obtain:

Dividing equation 5.14 by $${{e}^{-At}}$$, we obtain:

After some very simple manipulation we obtain:

Which, after applying the properties of exponents, can be rewritten as:

Given
and the following properties of the of the transition matrix $$\phi $$

Where $$I$$ is the identity matrix

Find
Equation 6.4 using equations 6.1, 6.2 and 6.3

Solution
Substituting $$x\left( t \right)$$ in the equation 6.1 by $$\underline{\phi }\left( t,{{t}_{0}} \right)$$ to obtain the differential equation of the state transition matrix, we get:

Re-arranging equation 6.5 in the following form:

It is clear that this equation complies with the first condition of exactness.

So to make this equation exact we will apply the Euler Integrating Factor Method.

It has been previously shown that in order to meet the second condition of exactness we can multiply equation 6.6 by a function h to obtain the following:

Now the second condition of exactness may be applied, where x is the dependent variable and t is the independent variable.

Now for the special case where $${{h}_{x}}$$ equals zero we obtain the following:

Which may be rearranged to obtain:

We can now solve for h as a function of t.

Substituting 6.12 in the equation 6.6 we get

We should recognize equation 6.13 as the derivative of $$\left( h\phi \right)$$, by using the relation given in equation 6.2:

Therefore, to find the solution we integrate both sides in the interval $${{t}_{o}}\le \tau \le t$$.

Equation 6.15 yields:

Re-arranging terms in equation 6.16 we obtain:

Substituting $$\underline{\phi }\left( t,{{t}_{0}} \right)$$ in the equation 6.17 by $$x\left( t \right)$$ we obtain the differential equation in the form of equation 6.4: