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[MA212/DE2] The Processes of solving systems of DEs, including homogeneous and Nonhomogeneous


Terms in this set (...)

Description of:
The method of solving a homogenous system of DEs, xvec' = A * xvec
1. Solve for eigenvalues/eigenvectors of A.
solve for det(A-lambda*I)=0, find lambda_1, etc...
Then solve for the augmented matrix (A-lambda_?*I | zero vector) (where ? is the ?th eigenvalue you found.)
(or solve systems of equations manually using the 1st, 2nd,...,last rows of A-lambda_?*I = 0 vector, but this way works easier for larger matrices)

If you get a parameter (or variable) when solving for the systems of eqns to get an eigenvector (ex. you get get infinitely many solutions and you get "t" as a parameter), then you should plug in a value for it to get the final v_?.

Each eigenvector is in the form
v_? =
x2], where x1 and x2 are the solutions to that system of equations.

2. Your general solution is then:
xvec = c1e^(lambda_1t)v1 + c2e^(lambda_2*t)*v2 + ... + cN*e^(lambda_Nt)e^(lambda_Nlambda_1*t)*v1 + c2*e^(lambda_2*t)*v2 + ... + cN*e^(lambda_N*t)*vN


c1,c2,...cN are the first,second,...last constants corresponding to the first,second,...,last (1st, 2nd,...,Nth) eigenvalue and first,second,...,last (1st, 2nd, ... Nth) eigenvector you found when solving for these

(Note: for constants, leave them as c1/c2/etc in gen. solution to DE. When solving an IVP, solve for these and put xvec=your vector and plug in the value of t into the equation),

and t is the variable in the solution to the DE for xvec.
Description of:
The method of undetermined coefficients to solve for a nonhomogenous system of DEs, xvec' = A * xvec + bvec
Also described: how to find a phase portrait from gen. soln.
1. solve associated homogeneous equation, e.g. xhvec' = A*xhvec (where A is a matrix that covers everything in xhvec...)

(see The method of solving a homogenous system of DEs, xvec' = A * xvec. Same thing except you solve for xhvec instead of xvec).

2. guess a possible xp (particular solution)
for example, if the final result depends on e^(5t). must have a constant matrix, always, no matter what!!!
xpvec = e^(5t)*
+ [b1
b2 ]
where a1, b1, a2, b2 are constants (b1/b2 have nothing to do with bvec in the original system, they are just constants).
If you have a 3x3 system of DEs
3. then solve xp' = A xp + bvec*. BE SURE TO TAKE THE DERIVATIVE OF xp, xp'. then solve for when all coefficients, a1, b1, a2, b2, etc... all equal 0 (solve systems of equations), to get the final values of all constants in xp. (maybe an equilibrium solution when all the coefficients equal 0, therefore those are solutions to the DE for all x and y (x vectors...)?)
4. FINALLY, your solution to the nonhomogeneous equation is: xvec = xhvec + xpvec
5. *note if you're asked to draw a phase portrait from that DE, you can easily use the eigenvalues from the xhvec part (like e^(lambda_1t)) in the general solution to the DE
Rule for sine and cosine in nonhomogeneous equations
If you have a sine or cosine in the nonhomogenous equation. you must have a sine and cosine in xpvec (because you will always get sine and cosine terms left over from taking the derivative).