Two entirely different topics here:

- How do we write out systems of differential equations and solve
them?

Let's step through an example slowly. We'll rewrite the differential equation

*x*'' + 2*x*' +*x*= sin(*t*)

*y*: let*y*=*x*'. Then, substituting*y*everywhere we see an*x*' in the original equation, we obtain the system of equations

*x*' =*y*

*y*' = -*x*- 2*y*+ sin(*t*)

*x*_{1},*x*_{2}, etc. In this case, we'd be taking*x*_{1}=*x*and*x*_{2}=*y*, to get the system

*x*_{1}' =*x*_{2}

*x*_{2}' = -*x*_{1}- 2*x*_{2}+ sin(*t*)

[ *x*_{1}'] = [ *x*_{2}]

*x*_{2}'- *x*_{1}- 2*x*_{2}+ sin(*t*)**[ ]**(square brackets) are the column vectors. So the column vector*x*' is the vector on the left-hand side, and that on the right-hand side is the column vector*f*(*t*,*x*_{1},*x*_{2}).Great! Now, what about solving these? For this problem we know how to solve it because we can solve the original second-order differential equation. The solution to this is

*x*=_{g}*x*+_{u}*x*, or_{d}

*x*= (_{g}*C*_{1}+*C*_{2}*t*)*e*- (1/2) cos(^{-t}*t*)

*x*was*x*_{1}and*x*' was*x*_{2}, so, taking the derivative of this, we have the*solution vector*

[ *x*_{1}] = [ ( *C*_{1}+*C*_{2}*t*)*e*- (1/2) cos(^{-t}*t*)

]

*x*_{2}(- *C*_{1}+ -*C*_{2}*t*+*C*_{2})*e*+ (1/2) sin(^{-t}*t*)

- How do we find the solution to a problem with sines and cosines
in the driving term?

For example, the previous problem (a coincidence?). Let's solve

*x*'' + 2*x*' +*x*= sin(*t*)

*x*'' + 2*x*' +*x*= 0, notice that this is the same as

*P*(D)[x] = (D + 1)^{2}[x] = 0

*x*=*e*, we get^{r t}

*P*(D)[*e*] = (D + 1)^{r t}^{2}[*e*] = 0^{r t}

*P*(D)[*e*] =^{r t}*e*^{r t}*P*(*r*), so

*e*(^{r t}*r*+ 1)^{2}= 0,

*r*= -1 (twice; it's a repeated root). So the undriven solution is

*x*= (_{u}*C*_{1}+*C*_{2}*t*)*e*^{-t}

Then, to find the driven solution we can proceed in one of two ways:

- Notice that the driving term
*f*(*t*) = sin(*t*) is the same as Imag(*e*). Thus, if we find a solution to^{i t}

*z*'' + 2*z*' +*z*=*e*,^{i t}

*x*, is just Imag(_{d}*z*).Let's do this. We want

*z*to be an exponential, so guess*z*=*A e*. Plugging in, we get^{i t}

*P*(D)[*z*] = (D + 1)^{2}[*z*] =*e*,^{i t}

(D + 1) Dividing by^{2}[*A e*] =^{i t}*e*, so^{i t}

*A e*(^{i t}*i*+ 1)^{2}=*e*.^{i t}

*e*and solving for^{i t}*A*, we get*A*= 1 / 2*i*. (Because (*i*+1)^{2}=*i*^{2}+ 2*i*+ 1 = -1 + 2*i*+ 1.) It's convenient to multiply the numerator and denominator of*A*by*i*to get*A*= -*i*/ 2. So

*z*= -(*i*/ 2)*e*= -(^{i t}*i*/ 2)(cos(*t*) +*i*sin(*t*)) = (1/2)(-*i*cos(*t*) + sin(*t*))

*x*= -(1/2) cos(_{d}*t*)

- Option two is to just guess

*x*=_{d}*A*cos(*t*) +*B*sin(*t*)

*A*and*B*work. Notice that we need to start with both the sine and the cosine term in most cases.The completion of this method is left as an exercise for the reader...

- Notice that the driving term

Gavin's DiffEq Clarification 000329

Last Modified: Wed Mar 29 17:44:51 CST 2000

Comments to glarose@umich.edu