r (t) = 2r(t) + sin t θ (t) = r(t) θ(t) + 1 = 1 1 θ(t) Write the given system in matrix form x = Ax + f ( ) sin(t) x y z = dy cos(t)

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1 Solutions HW Write the given system in matrix form x = Ax + f r (t) = 2r(t) + sin t θ (t) = r(t) θ(t) + We write this as ( ) r (t) θ (t) = ( ) ( ) 2 r(t) θ(t) + ( ) sin(t) Write the given system in matrix form x = Ax + f dx = x + y + z = 2x y + z dz = x + 5z dy We write this as dx dy dz = 2 x y 5 z Rewrite d y dy + y = cos(t) as a first order system in normal form. Note that the equation says that d y = dy y + cos(t). Setting x = y, x 2 = dy, x = d2 y, (so d y 2 = x 2 x + cos(t)) we get x x 2 = x = x x 2 x dy d 2 y 2 d y + = cos(t) x 2 x x 2 x + cos(t)

2 9.4. Write the given system as a set of scalar equations x = This becomes the equations ( ) ( ) 2 t x + e t x = 2x + x 2 + te t x 2 = x + x 2 + e t Determine whether the given vector functions are linearly dependent or independent on the interval (, ) ( ) ( ) sin t sin 2t, cos t cos 2t We compute the Wronskian ( ) sin t sin 2t det = sin t cos 2t sin 2t cos t = sin t cos t cos 2t where the last step can be deduced by using trig identities. Since sin t is not identically, the vector functions are linearly independent. (Alternatively, one can check that the Wronksian is nonzero at a point such as t = π 2.) Determine whether the given vector functions are linearly dependent or independent on the interval (, ), t t, t2 t 2 These functions are linearly independent, since a linear relations requires finding nonzero constants c, c 2, c such that c + c 2 t + c t 2 =. But, t, t 2 are linearly independent, so no such constants exist. Note that even though the vector functions are linearly independent, their Wronksian is still zero Determine whether the given functions form a fundamental solution set to an equation x (t) = Ax. If they do, find a fundamental matrix for the system and give a general solution.

3 x = et e t, x 2 = e t sin t cos t sin t We start by computing the Wronksian, x = cos t sin t cos t e t sin t cos t det e cos t sin t = e t (cos 2 t+sin 2 t) e t (sin t cos t sin t cos t)+e t (sin 2 t+cos 2 t) = 2e t e t sin t cos t Since this is nowhere, the solutions are linearly independent and form a fundamental set. A fundamental matrix is e t sin t cos t e cos t sin t e t sin t cos t and a general solution is c x + c 2 x 2 + c x Verify that the vector functions e t x =, x 2 = e t e t e t are solutions to the homogenous system on (, ) and that, x = 2 2 x = Ax = 2 2 x, 2 2 e t e t e t is a particular solution to 5t + x p = 2t 4t + 2 9t x = Ax + 8t = Ax + f(t)

4 Find a general solution to x = Ax + f(t). We check directly that e t e t x = = Ax e t x 2 = e t = Ax 2 e t x = e t = Ax e t 5 x p = 2 = Ax p + f(t) 4 A general solution to x = Ax + f(t) is c x + c 2 x 2 + c x + x p Prove that the operator L[x] = x Ax is a linear operator. We must show L[x + y] = L[x] + L[y] and L[cx] = cl[x]. L[x + y] = (x + y) A(x + y) = x + y Ax Ay = (x Ax) + (y Ay) = L[x] + L[y] L[cx] = (cx) A(cx) = cx cax = c(x Ax) = cl[x] Let X(t) be a fundamental matrix for the system x = Ax. Show that x(t) = X(t)X (t )x is the solution to the initial value problem x = Ax, x(t o ) = x. Since x(t) is a linear combination of the columns of the fundamental matrix, we just need to check that it satisfies the initial conditions. But x(t ) = X(t )X (t )x = Ix = x as desired, so x(t) is the dersired solutions Find eigenvalues and eigenvectors of the matrix We start by computing the characteristic polynomial.

5 det λ λ = λ + λ + 2 = (2 λ)( + λ) 2 λ So the eigenvalues are 2 and -. For λ = we must find the kernel of Row reducing we get which gives eigenvectors, For λ = 2 we must find the kernel of Row reducing we get which gives the eigenvector Find all eigenvalues and eigenvectors of

6 2 We start by computing the characteristic polynomial. λ 2 det λ = ( λ)(λ 2 2λ + 2) λ The first factor gives eigenvalue, the second gives eigenvalues ± i. For λ =, we must find the kernel of which gives the eigenvector 2 det For λ = i we must find the kernel of Solving this we get the eigenvector i 2 det i i 2 + i i Taking conjugates, we get that the eigenvector for λ = + i is 2 i i Find a general solution to the equation x = Ax where

7 A = 2 We start by computing the characteristic polynomial. λ det 2 λ = (λ 7λ 6) = (λ + )(λ + 2)(λ ) λ So the eigenvalues are, 2,. For λ =, we must find the kernel of Row reducing we get which gives the eigenvector For λ = 2, we must find the kernel of Row reducing we get which gives the eigenvector 4

8 For λ =, we must find the kernel of Row reducing we get which gives the eigenvector is 4 Combining these, we get that the general solution to the differential equation c e t + c 2 e 2t + c e t Find a fundamental matrix for the system x =Ax, where A = ( 5 ) 4 The characteristic polynomial of A is λ 2 5λ + 4 = (λ )(λ 4), so the eigenvalues are λ =, 4. For λ = we must find the kernel of ( ) which is spanned by ( )

9 For λ = 4 we must find the kernel of ( ) 4 which is spanned by 4 ( ) 4 The corresponding fundamental matrix is ( ) e t 4e 4t e t e 4t Find a general solution to the system of equations x = x 4y y = 4x 7y This system can be rewritten as x = Ax, where A = ( ) The characteristic polynomial is λ 2 + 4λ 5 = (λ )(λ + 5), so the eigenvalues are and -5. For λ = we must find the kernel of which is spanned by ( ) ( ) 2 For λ = 5 we must find the kernel of which is spanned by ( )

10 ( ) 2 Combining these, we see the general solution to the initial system is x = 2c e t + c 2 e 5t, y = c e t + 2c 2 e 5t Solve the initial value problem x (t) = x, x() = 4 From the eigenvectors and eigenvalues from problem 6, the general solution to this equation is x(t) = c e t + c 2 e t + c e 2t Plugging in the initial condition, we must solve the equations c c 2 = 4 c Row reducing the system and backsolving gives, c =, c 2 =, c =, so the desired solution is x(t) = e t e t + c e 2t a. Show that the matrix A = ( ) 4 has a repeated eigenvalue, and only one eigenvector. The characteristic polynomial is λ 2 +2λ+ = (λ+) 2, so the only eigenvalue is λ =. Searching for eigenvectors, we must find the kernel of ( ) 2 4 2

11 which is spanned by ( 2) b. Use your answer to part a. to find a nontrivial solution to x = Ax. ( ) e t 2 c. Try to find a second solution of the form te t u + e t u 2. Plugging this expression into x = Ax, we get te t u + e t u e t u 2 = te t Au + e t Au 2. Grouping the e t and te t terms together, we get to vector relations u 2 + u = Au 2 or (A + I)u 2 = u and u = Au, or (A + I)u =. We want u to be an eigenvector. To find u 2, we can either solve the given set of linear equations, or just guess a u 2 and see if (A + I)u 2 is an eigenvector. (This may seem ad hoc, but it works as long as your guess for u 2 is not already an eigenvector.) If we guess then ( u 2 = ) (A + I)u = ( ) = ( ) 2 4 which is an eigenvector. So we get a solution of the differential equation e t ( ) + te t ( 2 4 ) d. What is (A + I) 2 u 2? (A + I) 2 u 2 = (A + I)(A + I)u 2 = (A + I)u = from the equations we derived in part c Use the method of problem 5 to find a general solution to the system

12 x (t) = ( ) 5 x(t) Computing the characteristic polynomial, we get that λ = 2 is a double root, and ( v = ) is an eigenvector, so e 2t v is a solution to the differential equation. As in problem 5, we guess a solution of the form te 2t u + e 2t u 2. This gives rise to the equations (A 2I)u =, (A 2I) 2 u 2 = u. Guessing we get ( u 2 = ) ( ) ( ) u = = ( ) which is an eigenvector. So another solution is te 2t u + e 2t u 2. Combining these, we get a general solution ( ) ( ) ( ) c e 2t + c 2 (te 2t + e 2t )

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