MATH 2 INHOMOGENEOUS LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS W To find a particular solution for a linear inhomogeneous system of differential equations x Ax = ft) or of a mechanical system with external force ft) x Ax = ft) you can use the method of undetermined parameters Here is a short list of recommended guesses for various right hand sides ft) ey work almost always, but there are exceptional situations where they don t ese exceptions occur when the homogeneous equation x = Ax or x = Ax) has the recommended guess as solution When this happens multiplying the guess with t is o en an appropriate remedy ft) x p t) a p t k a P k t)p P k t) a polynomial of degree k e αt a e αt p sinαt)a + cosαt)b sinαt)p + cosαt)q e βt sinαt)a + e βt cosαt)b e βt sinαt)p + e βt cosαt)q In this table a, b are given constant vectors and α, β are given constant numbers, while p and q are constant vectors which play the role of the undetermined parameter Most of the time you will already have computed the eigenvalues and eigenvectors of the matrix A If you have a basis of eigenvectors of A which happens in all examples in the homework), then the computations become simpler ok, less complicated) if, instead of writing the undetermined vectors as p = [ you write them as linear combination of the eigenvectors us if your guess contains an undetermined vector p, then write it as p = p v + p 2 v 2 where v, v 2 are the eigenvectors of A, and solve for p and p 2 assuming A is a 2 2 matrix; if A is n n then you get n terms) e advantage of this is that it is now very easy to see the effect of multiplying p with A, namely, Ap = p Av + p 2 Av 2 = λ p v + λ 2 p 2 v 2
2 MATH 2 LINEAR INHOMOGENEOUS SYSTEMS a) b) Solve 2 P x = x + e αt, differential equation) 2 4 x) = initial condition) Note that the right hand side the so called forcing term ) has a parameter α in it We want to find the solution for all values of α 2 Outline of our computation Since the equation is of the form x = Ax + vft) the general solution is of the form x inh t) = x h t) + x p t) where x h t) is the general solution to the homogeneous equation and x p t) is a particular solution to the inhomogeneous equation We will first find x h t) by computing the eigenvalues&vectors of the matrix A e solutions will be of the form x h t) = c e λt v + c 2 e λ2t v 2, where Av = λ v and Av 2 = λ 2 v 2 Next, we find a particular solution by guessing Finally, we adjust the constants which appear in x h t) so as to match the initial conditions, ie we set t = in our general solution x inh t) and write out the equations x h ) + x p ) = is gives us two linear equations for the constants c, c 2 which appear in x h Once we have found those we are done 22 e homogeneous equation e homogeneous equation is x = Ax where the matrix A = [ 2 4 has eigenvalues and vectors [ [ λ = 2, v =, λ 2 = 5, v 2 = 2 erefore the solution to the homogeneous equation is 2) x h t) = c e 2t v + c 2 e 5t v 2 In our textbook this is called the complementary solution 2 A particular solution first attempt If you re in a hurry, skip this section and go to 4; this section is here because it follows the method suggested by the book, but 4 gives an easier way of finding the solution) Since the right hand side in the equation x Ax = ft) is given by ft) = e αt, ie since it is of the form e αt a constant vector, we look for a particular solution of the same form e simplest formula we could try is x p t) = e αt a = e αt a, a 2
MATH 2 LINEAR INHOMOGENEOUS SYSTEMS for some constant vector a If we substitute this in the le hand side of the equation x Ax = ft), then we will in the end) get a system of linear equation for a and a 2 : so x = αe αt a, Ax = A e αt a ) = e αt Aa, x Ax = αe αt a e αt Aa =? e αt Cancel the exponentials: αa Aa =, which leads to the following system of equations for a, a 2 α 2 α 4 Row reduction leads to messy algebra with α s: [ α 2α ) 2 2 2 α 4 2 α 4 α )R2+R [ 2 + α )α 4) 2 2 α 4 So we find that 2 a 2 = α 2 7α +, e particular solution is therefore x p t) = e αt a = a 2 a = α 4 2 a 2 = e αt α 2 7α + [ α 2 7α + 2 2 α 4 α 4 α 2 7α + α 4 2 A er this the computations get more and more complicated Instead of going on with this method we follow another approach which exploits the fact that we have already computed the eigenvalues&vectors of A 24 A particular solution second attempt We will still try a solution of the form xt) = e αt a, but instead of writing the unknown constant vector a as a column vector [ a a 2, we write it as a linear combination a = a v + a 2 v 2 of the two eigenvectors of the matrix A us we try a particular solution of the form x p t) = e αt{ a v + a 2 v 2 } = a e αt v + a 2 e αt v 2 To substitute this in the le hand side of the equation x Ax = f we compute and x pt) = αa e αt v + αa 2 e αt v 2 Ax p t) = a e αt Av + a 2 e αt Av 2 = a e αt λ v + a 2 e αt λ 2 v 2 = 2a e αt v + 5a 2 e αt v 2 where we have used that v, v 2 are eigenvectors of A with eigenvalues λ = 2, λ 2 = 5
4 MATH 2 LINEAR INHOMOGENEOUS SYSTEMS erefore ) x pt) Ax p t) = α 2)a e αt v + α 5)a 2 e αt v 2 To compare this expression with ft) we write ft) as a linear combination of the eigenvectors v and v 2 : e αt = ) v + ) v 2 We can do this by first writing [ as a combination of the eigenvectors, and then multiplying with e αt So first we find f and f 2 such that f v + f 2 v 2 = Since v = and v2 = [ 2 this leads to the following system of equations so that f 2 2 [ f R+R2 [ 2 / / f = 2, f 2 =, and hence = 2 v + v 2 erefore we can write the right hand side ft) of the differential equation as 4) ft) = 2 eαt v + eαt v By combining ) and 4) we see that x p Ax p = ft) will hold if α 2)a e αt v + α 5)a 2 e αt v 2 = 2 eαt v + eαt v holds Canceling the exponentials e αt on both sides gives us 5) α 2)a v + α 5)a 2 v 2 = 2 v + v 2 Since {v, v 2 } are linearly independent why?) equation 5) implies From here we find the coefficients a, a 2 : α 2)a = 2 and α 5)a 2 = a = 2/ α 2, a 2 = / α 5 e particular solution we get is e αt e αt 6) x p t) = 2 α 2 v + α 5 v 2 25 e general solution, and the one with the right initial value We can now get the general solution of the diffeq x Ax = f by adding the particular and the complementary solutions from 6) and 2) Since we have wri en both x p and x h as linear combinations of the eigenvectors, our formula for the general solution will also be such a combination Here it is: the general solution to a) is 7) xt) = c e 2t + 2 e αt ) v + c 2 e 5t + e αt ) α 2 v 2 α 5 To find the solution which also satisfies the initial condition b) we compute x) x) = c + 2 ) v + c 2 + ) α 2 v 2 α 5
MATH 2 LINEAR INHOMOGENEOUS SYSTEMS 5 erefore x) = holds exactly when c = 2 α 2 and c 2 = α 5 Substituting these values of c, c 2 in 7), we find that the solution to a) and b) is 8) xt) = 2 e αt e 2t α 2 v + e αt e 5t α 5 v 2 Find the solution of 9) x = A 4 x + 6 5 sin t, x) = Our plan We first find the eigenvectors/values of the matrix 4 A = 6 5 which appears in the problem this gives us the solution x h t) to the homogeneous equation en we compute a particular solution x p t) e general solution of the problem is then x g t) = x h t) + x p t) is general solution contains two constants By se ing t = in x g t) the initial condition gives us two equations for these constants, which we will then solve 2 e eigenvalues and vectors We have so the two eigenvalues are 2 4 deta λi) = λ 2 + 8λ 9 = λ + 9)λ ), λ =, λ 2 = 9 Solving A λi)v = for λ = λ, λ 2 leads us to the following eigenvectors [ 2 v =, v 2 = erefore the general solution to the homogeneous equation is ) x h t) = c e t v + c 2 e 9t v 2 Partial) Solution To find a particular solution for x Ax = [ sin t we recognize first that the right hand side has a trigonometric form sin t = sint) a + cost) b for certain constant vectors a, b In fact, a =, b = In this case you can almost) always find a particular solution of the same form, so we will set p sin t + q x p t) = sint) p + cost) q = cos t p 2 sin t + q 2 cos t
6 MATH 2 LINEAR INHOMOGENEOUS SYSTEMS and substitute in the equation Diligent computation then leads you to x q sin t + p p = cos t q 2 sin t + p 2 cos t p + 4p Ax p = 2 ) sin t + q + 4q 2 ) cos t 6p 5p 2 ) sin t + 6q 5q 2 ) cos t so that x p Ax p = p 4p 2 q ) sin t + p + q 4q 2 ) cos t 6p + 5p 2 q 2 ) sin t + p 2 6q + 5q 2 ) cos t e equation x p Ax p = [ sin t then leads to the following four equations for p, p 2, q, q 2 : p p 2 q q 2 4 4 6 5 6 5 4 Second approa use the eigenvectors of A Look for a particular solution in the form x p t) = c t)v + c 2 t)v 2, where v, v 2 are the eigenvectors of the matrix A above Since the right hand side of the equation contains the function sin t we will let the coefficients c t) and c 2 t) be similar trigonometric expressions erefore we will try x p t) = P sin t + Q cos t ) v + R sin t + S cos t ) v 2 Substituting in the le hand side of the equation leads to and x p = Q sin t + P cos t ) v + S sin t + R cos t ) v 2, x p t) = P sin t + Q cos t ) Av + R sin t + S cos t ) Av 2 = P sin t + Q cos t ) v + 9R sin t 9S cos t ) v 2, since Av = v and Av 2 = 9v 2 erefore ) x p Ax p = { Q P ) sin t + P Q) cos t } v On the other side of the equation x Ax = ft) we have sin t ft) = = sint) + { S + 9R) sin t + R + 9S) cos t } v 2 To compare this with our expression ) for x p Ax p, we write [ as a combination of the eigenvectors: = f v + f 2 v 2, Using v = [ and v 2 = [ 2 this leads to the equations [ f f 2 2 R+R2 [ 2 5 / 5 / 5
MATH 2 LINEAR INHOMOGENEOUS SYSTEMS 7 so that we have [ ft) = sint) = 5 sint)v + 5 sint)v 2 Comparing with ) we get the following equations for P, Q, R, S Solving these equations gives us Q P = 5 P Q =, S + 9R = 5 R + 9S = P = Q =, and R = 9 4, S = 4 We therefore get this particular solution: x p t) = sin t + cos t) v + 9 4 sin t 4 cos t) v 2 5 e general solution, and the one that satisfies the initial conditions Since the complementary solution we have found is a linear combination of v and v 2, it is easy to add x h and x p to get the general solution to the equation We get x g t) = x h t) + x p t) = c e t + sin t + cos t ) v + c 2 e 9t + 9 4 sin t 4 cos t )v 2 Finally, we can find the solution which satisfies the initial condition x) = [ 2 4 from 9) by computing To compare this with [ 2 4 leads to [ a a 2 2 2 4 x g ) = ) c + v + c 2 4) v2 we write this vector as a combination of the eigenvectors: [ 2 4 = a v + a 2 v 2 R+R2 [ 2 2 5 2 so that 2 = 4 4 5 v 2 5 v 2 e initial condition x g ) = [ 2 4 therefore implies c + = 4 5 and c 2 4 = 2 5, so c = c 2 = 6 4 e solution which satisfies the initial conditions is xt) = et + sin t + )v cos t + 2 4 / 5 2 / 5 6 4 e 9t + 9 4 sin t 4 cos t )v 2